[4 marks sum]

Question 14 Marks

How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78? Explain the double answer.

Answer

A.P. is $24,21,18, \ldots$Sum $=78$
Here, $a=24, d=21-24=-3$
$S _{ n }=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 78=\frac{n}{2}[2 \times 24+(n-1)(-3)]$
$\Rightarrow 156= n (48-3 n +3)$
$\Rightarrow 156=51 n-3 n^2$
$\Rightarrow 3 n^2-51 n+156=0$
$\Rightarrow 3 n^2-12 n-39 n+156=0 \quad \ldots\left\{\begin{array}{ll}\because 156 \times 3 & =468 \\ \therefore 468 & =-12 x-39 \\ -51 & =-12-39\end{array}\right\}$
$
\begin{aligned}
& \Rightarrow 3 n(n-4)-39(n-4)=0 \\
& \Rightarrow(n-4)(3 n-39)=0
\end{aligned}
$
Either $n-4=0$,
then $n=4$
or
$
3 n-39=0
$
then $3 n=39$
$
\Rightarrow n=13
$
$
\begin{aligned}
& \therefore n=4 \text { and } 13 \\
& n_4=a+(n-1) d \\
& =24+3(-3) \\
& =24-9 \\
& =15
\end{aligned}
$
$
\begin{aligned}
& n_{13}=24+12(-3) \\
& =24-36 \\
& =-12
\end{aligned}
$
$\therefore$ Sum of 5 th term to 13 term $=0$
$
(\because 12+9+6+3+0+(-3)+(-6)+(-9)+(-12)=0
$

View full question & answer→

Question 24 Marks

How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.

Answer

A.P. is $25,22,19, \ldots$
Sum $=116$
Here, $a=25, d=22-25=-3$
Let number of terms be $n$, then
$
\begin{aligned}
& 116=\frac{n}{2}[2 a+(n-1) d] \\
& \Rightarrow 232=n[2 \times 25+(n-1)(-3)] \\
& \Rightarrow 232=n[50-3 n+3] \\
& \Rightarrow n(53-3 n) \\
& \Rightarrow 232=53 n-3 n^2
\end{aligned}
$
$\Rightarrow 3 n^2-53 n+232=0 \quad \ldots\left\{\begin{array}{ll}\because 232 \times 3 & =696 \\ \therefore 696 & =-24 \times(-29) \\ -53 & =-24-29\end{array}\right\}$
$
\begin{aligned}
& \Rightarrow 3 n^2-24 n-29 n+232=0 \\
& \Rightarrow 3 n(n-8)-29(n-8)=0 \\
& \Rightarrow(n-8)(3 n-29)=0
\end{aligned}
$
Either $n-8=0$,
$
\text { then } n=8
$
or
$
3 n-29=0 \text {, }
$
then $3 n=29$
$
\Rightarrow n =\frac{29}{3}
$
which is not possible because of fraction
$
\therefore n =8
$
Now, $T=a+(n-1) d$
$
\begin{aligned}
& =25+7 \times(-3) \\
& =25-21 \\
& =4 .
\end{aligned}
$

View full question & answer→

Question 34 Marks

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.
Here,
The first term of the A.P (a) = 5
The last term of the A.P (l) = 45
Sum of all the terms Sn = 400
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,
$\begin{aligned} & 400=\left(\frac{n}{2}\right)(5+45) \\ & 400=\left(\frac{n}{2}\right)(50) \\ & 400=( n )(25) \\ & n =\frac{400}{25} \\ & n =16\end{aligned}$
Now, to find the common difference of the A.P. we use the following formula,
$
I=a+(n-1) d
$We get
$
\begin{aligned}
& 45=5+(16-1) d \\
& 45=5+(15) d \\
& 45=5=15 d \\
& \frac{45-5}{15}=d
\end{aligned}
$
Further, solving for $d$
$
\begin{aligned}
& d =\frac{40}{15} \\
& d =\frac{8}{3}
\end{aligned}
$
Therfore the number of terms is $n=16$ and the common difference of the A.p is $d=\frac{8}{3}$

View full question & answer→

Question 44 Marks

The sum of first six terms of an arithmetic progression is 42 . The ratio of the 10 th term to the 30 th term is $\frac{1}{3}$. Calculate the first and the thirteenth term.

Answer

$
T_{10}: T_{30}=1: 3, S_6=42
$
Let $a$ be the first term and $d$ be a common difference, then
$
\begin{aligned}
& \frac{a+9 d}{a+29 d}=\frac{1}{3} \\
& \Rightarrow 3 a+27 d=a+29 d \\
& \Rightarrow 3 a-a=29 d-27 d \\
& \Rightarrow 2 a=2 d \\
& \Rightarrow a=d
\end{aligned}
$
Now, $S _6=42$
$
\begin{aligned}
& =\frac{n}{2}[2 a+(n-1) d] \\
& \Rightarrow 42=\frac{6}{2}[2 a+(6-1) d] \\
& \Rightarrow 42=3[2 a+5 d] \\
& \Rightarrow 14=2 a+5 d \\
& \Rightarrow 14=2 a+5 a \quad \ldots(\because d=a) \\
& \Rightarrow 7 a=14 \\
& \Rightarrow a=\frac{14}{7}=2 \\
& \therefore a=d=2
\end{aligned}
$
Now $_x T_{13}=a+(n-1) d$
$
\begin{aligned}
& =2+(13-1) \times 2 \\
& =2+12 \times 2 \\
& =2+24 \\
& =26
\end{aligned}
$
$\therefore 1$ st term is 2 and thirteenth term is 26 .

View full question & answer→

Question 54 Marks

The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 equal to the sum of first 2n terms of another A.P. whose first term is – 30 and the common difference is 8. Find n.

Answer

$
\begin{aligned}
& \text { In an A.P. } \\
& S_n=S_{2 n}
\end{aligned}
$
For the first A.P. $a=8, d=20$
and for second A.P. $a=-30, d=8$
Now, $S _{ n }=\frac{n}{2}[2 a+(n-1) d]$
$
\begin{aligned}
& =\frac{n}{2}[2 \times 8+(n-1) \times 20] \\
& =\frac{n}{2}[16+20 n-20] \\
& =\frac{n}{2}[20 n-4] \\
& =10 n^2-2 n......(1)
\end{aligned}
$
Similarly,
$
\begin{aligned}
& S_{2 n}=2 \frac{n}{2}[2 a+(2 n-1) d] \\
& = n [2 \times(-30)+(2 n-) \times 8] \\
& = n [-60+16 n -8] \\
& = n (16 n -68) \\
& =16 n ^2-68 n \\
& \because S_n=22 n \\
& \therefore 10 n^2-2 n =16 n ^2-68 n \\
& \Rightarrow 16 n^2-10 n^2-68 n +2 n =0 \\
& \Rightarrow 6 n^2-66 n =0 \\
& \Rightarrow n^2-11 n =0 \\
& n ( n -11)=0
\end{aligned}
$
Either $n=0$ which is not possible or
$
n-11=0 \text {, }
$
then $n =11$
$
\therefore n =11 \text {. }
$

View full question & answer→

Question 64 Marks

If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Answer

Let $a$ and $d$ be the first term and common difference, respectively of an AP.
$\because$ Sum of $n$ terms of an AP
$
S_n=\frac{n}{2}[2 a+(n-1) d].....(1)
$
Now, $S_6=36$[Given]
$
\begin{aligned}
& \Rightarrow \frac{6}{2}[2 a+(6-1) d]=36 \\
& \Rightarrow 2 a+5 d=12 \ldots . . \text { (2) }
\end{aligned}
$
And $S_{16}=256$
$
\begin{aligned}
& \Rightarrow \frac{16}{2}[2 a+(16-1) d]=256 \\
& \Rightarrow 2 a+15 d=32 \ldots \ldots \text { (3) }
\end{aligned}
$
On subtracting equation (2) from equation (3), we get
$
\begin{aligned}
& 10 d=20 \\
& \Rightarrow d=2
\end{aligned}
$
From equation (2)
$
\begin{aligned}
& 2 a+5(2)=12 \\
& \Rightarrow 2 a=12-10=2 \\
& \Rightarrow a=1 \\
& \therefore S_{10}=\frac{10}{2}[2 a+(10-1) d] \\
& =5[2(1)+9(2)] \\
& =5(2+18) \\
& =5 \times 20 \\
& =100
\end{aligned}
$
Hence, the required sum of first 10 terms is 100.

View full question & answer→

Question 74 Marks

The sum of 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find the A.P.

Answer

Let $a$ be the first term and $d$ be the common difference, then
$
\begin{aligned}
& a_5=a+(5-1) d=a+4 d \\
& a_7=a+(7-1) d=a+6 d \\
& a_5+a_7=a+4 d+a+6 d=52 \\
& \Rightarrow 2 a+10 d=52 \\
& \Rightarrow a+5 d=26......(1)
\end{aligned}
$
Similarly,
$
\begin{aligned}
& a_{10}=a+(10-1) d \\
& =a+9 d \\
& \Rightarrow a+9 d=46......(2)
\end{aligned}
$
Subtracting (1) from (2),
$
\begin{aligned}
& 4 d=20 \\
& \Rightarrow d=\frac{20}{4}=5 \\
& \Rightarrow d=5
\end{aligned}
$
Now, put the value of $d$ in eq. (1)
$
\begin{aligned}
& a+5 \times 5=26 \\
& \Rightarrow a=26-25 \\
& \Rightarrow a=1
\end{aligned}
$
Hence,
$
\begin{aligned}
& a_2=a_1+d \\
& =1+5 \\
& =6
\end{aligned}
$
$
\begin{aligned}
& a_3=a_2+d \\
& =6+5 \\
& =11 \\
& a_4=a_3+d \\
& =11+5 \\
& =16
\end{aligned}
$
$\therefore$ The A.P formed is $1,6,11,16, \ldots$

View full question & answer→

Question 84 Marks

If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its 63 rd term.

Answer

$\begin{aligned} & a_7=\frac{1}{9} \\ & \Rightarrow a+6 d=\frac{1}{9} \ldots(i) \\ & a_9=\frac{1}{7} \\ & \Rightarrow a+8 d=\frac{1}{7} \ldots \ldots \text { (ii) } \\ & a_7=\frac{1}{9} \Rightarrow a+6 d=\frac{1}{9} \\ & a_9=\frac{1}{7} \Rightarrow a+8 d=\frac{1}{7}\end{aligned}$

$
\begin{aligned}
& -2 d=\frac{7-9}{63} \\
& -2 d=\frac{-2}{63} \\
& d=\frac{1}{63}
\end{aligned}
$
Now, substitute the value of $d$ in eq. (i), we get
$
\begin{aligned}
& a+6\left(\frac{1}{63}\right)=\frac{1}{9} \\
& a=\frac{1}{9}-\frac{6}{63} \\
& =\frac{7-6}{63} \\
& =\frac{1}{63} \\
& \therefore a_{63}=a+62 d \\
& =\frac{1}{63}+62\left(\frac{1}{63}\right) \\
& =\frac{1+62}{63} \\
& =\frac{63}{63} \\
& =1 .
\end{aligned}
$

View full question & answer→

Question 94 Marks

Find the $31^{\text {st }}$ term of an $A.P$. whose $11^{\text {th }}$ term is $38$ and $6^{\text {th }}$ term is $73 .$

Answer

Given: $t _{11}=38, t _6=73$
To Find: $t _{31}$
Solution:
Let $a$ be the first term and $d$ be the common difference.
$t _{11}=38$
$\therefore t _{11}= a +( n -1) d$
$\therefore t _{11}= a +(11-1) d$
$\therefore 38= a +10 d \ldots \text { (i) }$
Similarly, in the case of $t_6=73$
$\therefore 73=a+5 d......(ii)$
Subtracting equation $(i)$ from $(ii)$
$73=a+5 d$
$-38=a+10 d$
$(-)(-)(-)$
$\therefore 35=-5 d$
$\therefore d=\frac{35}{-5}$
$\therefore d=-7$
Substitute the value of $d$ in equation $(i),$ we get,
$a+10 d=38$
$a+10(-7)=38 \ldots(d=-7)$
$a-70=38$
$a=38+70$
$a=108$
$\therefore t_{31}=a+(31-1) d$
$\therefore t_{31}=a+30 d$
$\therefore t_{31}=108+30(-7) \ldots(a=108, d=-7)$
$\therefore t_{31}=108-210$
$\therefore t_{31}=-102$
$\therefore 31^{\text {th }} \text { term }=-102 .$

View full question & answer→

Mathematics [4 marks sum]

Test yourself on this topic