[3 marks sum] - Mathematics STD 10 Questions - Vidyadip

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[3 marks sum]

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Question 13 Marks

Solve for x : 1 + 4 + 7 + 10 + … + x = 287.

Answer

$
1+4+7+10+\ldots+x=287
$
Here, $a=1, d=4-1=3, n=x$
$I=x=a=(n-1) d=1+(n-1) \times 3$
$\Rightarrow x -1=( n -1) d$
$S _{ n }=\frac{n}{2}[2 a+(n-1) d]$
$287=\frac{n}{2}[2 \times 1+(n-1) 3]$
$574=n(2-3 n-3)$
$\Rightarrow 3 n^2-n-574=0$
$\Rightarrow 3 n^2-42 n+41 n-574=0$
$\Rightarrow 3 n ( n -14)+41( n -14)=0$
$\Rightarrow(n-14)(3 n+41)=0$
Either $n-14=0$,
then $n=14$
or
$
3 n+41=0 \text {, }
$
then $3 n=-41$
$
\Rightarrow n=\frac{-41}{3}
$
Which is not possible being negative.
$
\therefore n=14
$
Now, $x = a +( n -1) d$
$
\begin{aligned}
& =1+(14-1) \times 3 \\
& =1+13 \times 3 \\
& =1+39 \\
& =40 \\
& \therefore x=40 .
\end{aligned}
$

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Question 23 Marks

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

$
\begin{aligned}
& \text { First term of an A.P. }(a)=17 \\
& \text { and last term }(l)=350 \\
& d=9 \\
& =T_n=a+(n-1) d \\
& 350=17+(n-1) \times 9 \\
& \Rightarrow 350-17=9(n-1) \\
& \Rightarrow 333=9(n-1) \\
& \Rightarrow n-1=\frac{33}{9}=37 \\
& n=37+1=38
\end{aligned}
$
and
$
\begin{aligned}
& \text { Sn }=\frac{n}{2}[2 a+(n-1) d] \\
& =\frac{38}{2}[2 \times 17+(38-1) \times 9] \\
& =19[34+37 \times 9] \\
& =19[34+333] \\
& =19 \times 367 \\
& =6973
\end{aligned}
$
Hence $n=38$ and $S_n=6973$.

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Question 33 Marks

The sum of first $15$ terms of an $A.P.$ is $750$ and its first term is $15$. Find its $20^{th}$ term.

Answer

Let $a$ be the first term and $d$ be the common difference.
Now $, a =15$
Sum of first $n$ terms of an AP is given by,
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{15}=\frac{15}{2}[2 a+(15-1) d]$
$\Rightarrow 750=\frac{15}{2}(2 a+14 d)$
$\Rightarrow a+7 d=50$
$\Rightarrow 15+7 d=50$
$\Rightarrow 7 d=35$
$\Rightarrow d=5$
Now, $20^{\text {th }}$ term $=a_{20}$
$ =a+19 d$
$=15+19 \times 5$
$=15+95$
$=110 $

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Question 43 Marks

In an $A.P.$ (with usual notations) : given $a = 8, a_n = 62, S_n = 210$, find n and d

Answer

$ a=8, a_n=62, S_n=210$
$a_n=a+(n-1) d$
$62=8+(n-1) d$
$(n-1) d=62-8=54\ ...(i)$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$210=\frac{n}{2}[2 \times 8+54]......(from 1)\ $
$420=n(16+54)$
$\Rightarrow 420=70 n$
$n=\frac{420}{70}=6$
$\therefore(6-1) d=54$
$\Rightarrow 5 d=54$
$\Rightarrow d=\frac{54}{5}$
Hence $d =\frac{54}{5}$ and $n=6$.

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Question 53 Marks

In an A.P. (with usual notations) : given $d=5, S_9=75$, find a and $a _9$

Answer

$d =5, S _9=75$
$a _{ n }= a +( n -1) d$
$a _9= a +(9-1) \times 5$
$= a +40...(i)$
$S _9=\frac{n}{2}[2 a+(n-1) d]$
$75=\frac{9}{2}[2 a+8 \times 5]$
$\frac{150}{9}=2 a +40$
$2 a =\frac{150}{9}-40$
$=\frac{50}{3}-40$
$2 a =\frac{-70}{3}$
$\Rightarrow a =\frac{-70}{2 \times 3}$
$a =\frac{-35}{3}$
From (i),
$a g=a+40$
$=\frac{-35}{3}+40$
$=\frac{-35+120}{3}$
$=\frac{85}{3}$
$\therefore a=\frac{-35}{3}, a_9=\frac{85}{3}$.

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Question 63 Marks

In an $A.P.$ (with usual notations) : given $a = 7, a_{13} = 35$, find d and $S_{13}$

Answer

$
\begin{aligned}
& a=7, a_{13}=35 \\
& a_n=a+(n-1) d \\
& 35=7+(13-1) d \\
& \Rightarrow 35-7=12 d \\
& \Rightarrow 28=12 d \\
& \Rightarrow d=\frac{28}{12} \\
& =\frac{7}{3} \\
& =2 \frac{1}{3}
\end{aligned}
$
and
$
\begin{aligned}
& S_{13}=\frac{n}{2}[2 a+(n-1) d] \\
& =\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right] \\
& =\frac{13}{2}[14+28] \\
& =\frac{13}{2} \times(42) \\
& =13 \times 21 \\
& =273 .
\end{aligned}
$

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Question 73 Marks

In an A.P. (with usual notations) : given $a =5, d=3, a _{ n }=50$, find n and $S _{ n }$

Answer

$a=5, d=3, a_n=50$
$a_n=a+(n-1) d$
$50=5+(n-1) \times 3$
$\Rightarrow 50-5=3(n-1)$
$\Rightarrow 45=3(n-1)$
$\Rightarrow \frac{45}{3}=n-1$
$\Rightarrow n-1=15$
$\Rightarrow n=15+1=16$
$\therefore n=16$
and
$ S_n=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{16}{2}[2 \times 5+(16-1) \times 3]$
$=8[10+45]$
$=8 \times 55$
$=440 .$

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Question 83 Marks

Find the sums given below : – 5 + ( – 8) + ( – 11) + … + ( – 230)

Answer

$
-5+(-8)+(-11)+\ldots+(-230)
$
Here, $a=-5, d=-8-(-5)=-8+5=3$
$
\begin{aligned}
& I=-230 \\
& \therefore I=a+(n-1) d \\
& \Rightarrow-230=-5+(n-1)(-3) \\
& -230+5=-3(n-1) \\
& \Rightarrow-225=-3(n-1) \\
& \frac{-225}{-3}=n-1 \\
& \Rightarrow n-1=75 \\
& \Rightarrow n=75+1=76 \\
& \therefore S_n=\frac{n}{2}[a+l] \\
& =\frac{76}{2}[-5+(-230)] \\
& =38[-5-230] \\
& =38 \times(-235) \\
& =-8930 .
\end{aligned}
$

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Question 93 Marks

Find the sums given below : 34 + 32 + 30 + … + 10

Answer

$\begin{aligned} & 34+32+30+\ldots+10 \\ & \text { Here, } a=34, d=32-34=-2, I=10 \\ & T_n=a+(n-1) d \\ & 10=34+(n-1)(-2) \\ & -24=-2(n-1) \\ & =\frac{-24}{-2} \\ & =n-1 \\ & =12 \\ & \therefore n=12+1=13 \\ & S_n=\frac{n}{2}[a+l] \\ & =\frac{13}{2}[34+10] \\ & =\frac{13}{2} \times 44 \\ & =286 .\end{aligned}$

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Question 103 Marks

How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero?

Answer

$
\begin{aligned}
& \text { A.P. }=27,24,21, \ldots \\
& a=27 \\
& d=24-27=-3 \\
& S_n=0
\end{aligned}
$
Let $n$ terms be there in A.P.
$
\begin{aligned}
& S_n=\frac{n}{2}[2 a+(n-1) d] \\
& \Rightarrow 0=\frac{n}{2}[(2 \times 27)+(n-1)(-3)] \\
& \Rightarrow 0=n[54-3 n+3] \\
& \Rightarrow n[57-3 n]=0 \\
& \Rightarrow(57-3 n)=\frac{0}{n}=0 \\
& \Rightarrow 3 n=57 \\
& \therefore n=\frac{57}{3} \\
& =19 .
\end{aligned}
$

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Question 113 Marks

Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.

Answer

Let $a$ and $d$ be the first term and common difference of A.P. respectively.
Given, $a =5$
$
\begin{aligned}
& a_1+a_2+a_3+a_4=\frac{1}{2}\left(a_5+a_6+a_7+a_8\right) \\
& \therefore a+(a+d)+(a+2 d)+(a+3 d) \\
& =\frac{1}{2}[(a+4 d)+(a+5 d)+(a+6 d)+(a+7 d)] \\
& \Rightarrow 2(4 a+6 d)=(4 a+22 d) \\
& \Rightarrow 2(20+6 d)=(20+22 d) \quad \ldots(\because a=5) \\
& \Rightarrow 40+12 d=20+22 d \\
& \Rightarrow 10 d=20 \\
& \Rightarrow d=2
\end{aligned}
$
Thus, the common difference of A.P. is 2 .

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Question 123 Marks

If $a_n=3-4 n$, show that $a_1, a_2, a_3, \ldots$ form an A.P. Also find $S_{20}$.

Answer

$
\begin{aligned}
& a_n=3-4 n \\
& a_1=3-4 \times 1=3-4=-1 \\
& a_2=3-4 \times 2=3-8=-5 \\
& a_3=3-4 \times 3=3-12=-9 \\
& a_4=3-4 \times 4=3-16=-13 \text { and so on }
\end{aligned}
$
Here, $a=-1, d=-5-(-1)=-5+1=-4$
Now $_1 S_{20}=\frac{n}{2}[2 a+(n-1) d]$
$
\begin{aligned}
& =\frac{20}{2}[2 \times(-1)+(20-1) \times(-4)] \\
& =10[-2+19 \times(-4)] \\
& =10[-2-76] \\
& =10 \times(-78) \\
& =-780 .
\end{aligned}
$

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Question 133 Marks

Show that $a_1, a_2, a_3, \ldots$ form an $A.P$. where $a_n$ is defined as $a_n=3+4 n$. Also find the sum of first $15$ terms.

Answer

$a_n=3+4 n$
$a_1=3+4 \times 1=3+4=7$
$a_2=3+4 \times 2=3+8=11$
$a_3=3+4 \times 3=3+12=15$
$a_4=3+4 \times 4=3+16=19$
and so on Here, $a=1$ and $d=11-7=4$
$ S_{15}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{15}{2}[2 \times 7+(15-1) \times 4]$
$=\frac{15}{2}[14+14 \times 14]$
$=\frac{15}{2}[14+56]$
$=\frac{15}{2} \times 70$
$=525 $

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Question 143 Marks

Find the sum of the following A.P.s: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms

Answer

$
\begin{aligned}
& \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots \text { to } 11 \text { terms } \\
& a=\frac{1}{15} \\
& d=\frac{1}{12}-\frac{1}{15} \\
& =\frac{5-4}{60}=\frac{1}{60}
\end{aligned}
$
$
\begin{aligned}
& \text { or } \\
& \frac{1}{10}-\frac{1}{12} \\
& =\frac{6-5}{60}=\frac{1}{60} \\
& n=11 \\
& \therefore S _{11}=\frac{n}{2} \times[2 a+(n-1) d] \\
& =\frac{11}{2} \times\left[2 \times \frac{1}{15}+(11-1) \times \frac{1}{60}\right] \\
& =\frac{11}{2} \times\left[\frac{2}{15}+\frac{1}{6}\right] \\
& =\frac{11}{2} \times\left[\frac{4+5}{30}\right] \\
& =\frac{11}{2} \times \frac{9}{30} \\
& =\frac{33}{20}
\end{aligned}
$
or
$
=1 \frac{13}{20} \text {. }
$

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Question 153 Marks

If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.

Answer

$
\begin{aligned}
T_3=1, & T_6=-11, n=32 \\
a & +2 d=1......(1) \\
a & +5 d=-11.....(2) \\
&- -\quad+
\end{aligned}
$
Subtracting (1) and (2),
$
\begin{aligned}
& -3 d=12 \\
& \Rightarrow d=\frac{12}{-3}=-4
\end{aligned}
$
Substitute the value of $d$ in eq. (1)
$
\begin{aligned}
& a+2(-4)=1 \\
& \Rightarrow a-8=1 \\
& a=1+8=9 \\
& \therefore a=9, d=-4 \\
& S_{32}=\frac{n}{2}[2 a+(n-1) d] \\
& =\frac{32}{2}[2 \times 9+(32-1) \times(-4)] \\
& =16[18+31 \times(-4)] \\
& =16[18-124] \\
& =16 \times(-106) \\
& =-1696
\end{aligned}
$

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Question 163 Marks

Find the sum of the two middle most terms of the A.P. $-\frac{4}{3},-1,-\frac{2}{3}, \ldots, 4 \frac{1}{3}$

Answer

$
\text { A.P. }-\frac{4}{3},-1,-\frac{2}{3}, \ldots, 4 \frac{1}{3}
$
Here, $a=-\frac{4}{3}, d=-1-\left(\frac{-4}{3}\right)-1+\frac{4}{3}=\frac{1}{3}$
$
\begin{aligned}
& I=4 \frac{1}{3} \\
& \therefore T _{ n }= I =4 \frac{1}{3}=a+(n-1) d \\
& \Rightarrow 4 \frac{1}{3}=\frac{-4}{3}+(n-1) \times \frac{1}{3} \\
& \therefore \frac{13}{3}+\frac{4}{3}=\frac{1}{3}(n-1) \\
& \Rightarrow \frac{17}{3} \times \frac{3}{1}=( n -1) \\
& n -1=17 \\
& \Rightarrow n =17+1 \\
& =18
\end{aligned}
$
$\therefore$ Two middle term are $\frac{18}{2}$ and $\frac{18}{2}+1$
$=9$ th and 10th term
$
\begin{aligned}
& \therefore a_9+a_{10}=a+8 d+a+9 d \\
& =2 a+17 d \\
& =2 \times\left(\frac{-4}{3}\right)+17 \times \frac{1}{3} \\
& =\frac{-8}{3}+\frac{17}{3} \\
& =\frac{9}{3} \\
& =3 .
\end{aligned}
$

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Question 173 Marks

Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.

Answer

A.P. is $3,8,13, \ldots, 253$
12th term from the end
Last term $=253$
Here, $a=3, d=8-3=5$
$\therefore$ Last term $(n)=a+(n-1) d$
$253=3+(n-1) \times 5$
$\Rightarrow 253=3+5 n -5$
$\Rightarrow 253-3+5=5 n$
$\Rightarrow 5 n =225$
$\Rightarrow n =\frac{255}{5}=51$
$\therefore 253$ is 51 th term
Let $m$ be the 20th term from the last term
Then $m$ be the 20th term from the last term
Then $m=1-(n-1) d$
$=253-(20-1) \times 5$
$=253-19 \times 5$
$=253-95$
$=158$
$\therefore 20$ th term from the end $=158$.

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Question 183 Marks

The sum of the first three terms of an A.P.is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.

Answer

Let the three numbers in A.P. are
$
a-d_1 a, a+d
$
Now, $a-d+a+a+d=33$
$
\begin{aligned}
& \Rightarrow 3 a=33 \\
& \Rightarrow a=\frac{33}{3}=11
\end{aligned}
$
And $(a-d)(a+d)=a+29$
$a^2-d^2=a+29$
$(11)^2-d^2=11+29$
$\Rightarrow 121-d^2=40$
$d^2=121-40$
$=81$
$=( \pm 9)^2$
$\therefore d= \pm 9$
If $d=9$, then
$\therefore$ Numbers are $11-9,11,11+9$
$\Rightarrow 2,11,20$
If $d=-9$, then
$11+9,11,11-9$
$\Rightarrow 20,11,2$
Hence numbers are $2,11,20$ or $20,11,2$.

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Question 193 Marks

The sum of three numbers in A.P. is 30 and the ratio of first number to the third number is 3 : 7. Find the numbers.

Answer

Sum of three numbers in A.P. $=30$
Ratio between first and the third number $=3: 7$
Let numbers be
$a-d_t a_i a+d_1$ then
$a-d+a+a+d=30$
$\Rightarrow 3 a =30$
$\Rightarrow a=\frac{30}{3}=10$
and
$\frac{a-d}{a+d}=\frac{3}{7}$
$\Rightarrow 7 a -7 a =3 a +3 d$
$\Rightarrow 7 a -3 a =3 d +7 d$
$\Rightarrow 4 a =10 d$
$\Rightarrow 10 d =4 \times 10=40$
$\Rightarrow d=\frac{40}{10}=4$
$\therefore$ Numbers are $10-4,10,10+4$
$\Rightarrow 6,10,14$.

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Question 203 Marks

The sum of three numbers in A.P. is 3 and their product is – 35. Find the numbers.

Answer

Sum of three numbers which are in A.P. $=3$
Their product $=-35$
Let three numbers which are in A.P.
$
\begin{aligned}
& a-d_x a_t a+d \\
& a-d+a+a+d=3 \\
& \Rightarrow 3 a=3 \\
& \Rightarrow a=\frac{3}{3}=1
\end{aligned}
$
and
$(a-d) x a(a+d)=-35$
$(1-d) \times 1 \times(1+d)=-35$
$1^2-d^2=-35$
$1-d^2=-35$
$\Rightarrow d^2=35+1=36$
$\therefore d= \pm 6$
If $d=6$
$\therefore$ Number are $1-6,1,1+6$
$=-5,1,7$
If $d=-6$
$1+6,1,1-6$
$\Rightarrow 7,1,-5$
Hence numbers on A.P. are
$
-5,1,7 \text { or } 7,1,-5
$

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Question 213 Marks

If the $n$th terms of the two A.g.s $9,7,5, \ldots$ and $24,21,18, \ldots$ are the same, find the value of $n$. Also, find that term

Answer

nth term of two A.P.s $9, 7, 5,…$ and $24, 21, 18, …$ are same
In the first $A.P. 9, 7, 5, …$
$a = 9$ and $d = 7 – 9 = –2$
$T_n = a + (n – 1)d$
$= 9 + (n – 1)(–2)$
$= 9 – 2n + 2$
$= 11 – 2n$
and in second $A.P. 24, 21, 18,...$
$a_1 = 24, d_1 = 21 – 24 = –3$
$T_n = 24 + (n – 1)(–3)$
$= 24 – 3n + 3$
$= 27 – 3n$
$\because $ The nth terms of both A.P. is same
$\therefore 11 – 2n = 27 – 3n$
$–2n + 3n = 27 – 11$
$\Rightarrow n = 16$
and
$T_{16} = a + (n – 1)d$
$= 9 + 15 x (–2)$
$= 9 – 30$
$= –21.$

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Question 223 Marks

Which term of the A.P. 3, 10, 17,… will be 84 more than its 13th term?

Answer

A.P. is $3,10,17, \ldots$
Here, $a=3, d-10-3=7$
$
\begin{aligned}
& T _{13}= a +12 d \\
& =3+12 \times 7 \\
& =3+84 \\
& =87
\end{aligned}
$
Let $n$th term is 84 more then its 13 th term
$
\begin{aligned}
& \therefore T_n=84+87=171 \\
& \Rightarrow a+(n-1) d=171 \\
& \Rightarrow 3+) n-1) \times 7=171 \\
& (n-1) \times 7 \\
& =171-3 \\
& =168 \\
& n-1=\frac{168}{7}=24 \\
& n=24+1 \\
& =25
\end{aligned}
$
$\therefore 25$ th term is the required term.

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Question 233 Marks

The sum of 2nd and 7th terms of an A.P. is 30. If its 15th term is 1 less than twice its 8th term, find the A.P.

Answer

Addition of and term of AP = 30
(A+ D) + (A + 6D) = 30
Sum = 2A + 7D = 30.
Now as the question says the subtraction of 15th term and 2(8th term) = 1, that will be
(A + 14D) – 2(A + 7D) = 1
(A + 14D) – 2(A + 7D) = 1
A – 2A = 1
Value of A = 1
First term = 1,
Difference can be found by substituting value of A in (A+ D) + (A + 6D) = 30 ) we get,
(1+ D) + (1 + 6D) = 30
2 + 7D = 30
D = 4.
With both the values of A and D, the Arithmetic Progression as 1, 5, 9, 13…..

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Question 243 Marks

The 15th term of an $A.P.$ is $3$ more than twice its $7th$ term. If the 10th term of the $A.P.$ is $41$, find its nth term.

Answer

Let a be the first term and d be a common difference.
We have,
$a_{10} = 41$
$\Rightarrow a + 9d = 41 ...(i)$
and
$a_{15} = 2a + 3$
$\Rightarrow a + 14d = 2(a + 6d) + 3$
$\Rightarrow a + 14d = 2a + 12d + 3$
$\Rightarrow a – 2d = 3 ...(ii)$
Subtracting $(ii)$ from $(i)$, we get
$9d + 2d = 41 + 3$
$\Rightarrow 11d = 44$
$\Rightarrow d = 4$
Now, from (i), we get
$a + 9 x 4 = 41$
$\Rightarrow a + 36 = 41$
$\Rightarrow a = 5$
Now,
nth term
$= a_n$
$= a + (n – 1)d$
$= 5 + (n – 1)4$
$= 4n + 1.$

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[3 marks sum] - Mathematics STD 10 Questions - Vidyadip