Download App

P-1 Arithmetic Progression — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsP-1 Arithmetic Progression2 Marks
Question
Write the correct number in the given boxes from the following A. P.
3, 6, 9, 12, . . .
$\text { Here } t _1=\square, t _2=\square, t _3=\square, t _4=\square, $
$ t _2- t _1=\square, t _3- t _2=\square$
$\therefore d =\square$
✓

Answer

$3,6,9,12, \ldots$
First term a $=3$
Second term $t _1=6$
Third term $t_2=9$
Fourth term $t_3=12$
We know that $d=t_{n+1}-t_n$
Thus, $t_2-t_1=9-6=3$
$t_3-t_2=12-9=3$
Thus, $d=3$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Complete the following activities and rewrite it : (2M)" from P-1 Arithmetic Progression→P-1 Arithmetic Progression — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Similar questions

Complete the following activity to find the 27 th term or the A.P. 9. 4. -1,-6,-11, ........
Image
→
If α, β are roots of quadratic equation,
→
Draw a circle with center O and radius 3 cm
↓
Take any point P on the circle
↓
Draw ray OP
↓
Draw perpendicular to ray OP from point P
→
First term and common difference of an A.P. are 6 and 3 respectively ; find $S_{27}$.
$a=6, d=3, s_{27}=?$
$ S _{ n }  =\frac{ n }{2}[\square+( n -1) d ] $
$ S _{27}  =\frac{27}{2}[12+(27-1) \square] $
$ =\frac{27}{2} \times \square $
$ =27 \times 45=\square$
→
In $\triangle \mathrm{ABC}$, ray $\mathrm{BD}$ bisects $\angle \mathrm{ABC}$. $A-D-C$, side DE $\|$ side $B C, A-E-B$ then prove that, $\frac{A B}{B C}=\frac{A E}{E B}$
→
Complete the following activity to find the number of natural numbers from 1 to 171 which are divisible by 5 .
Image
→
→
Complete the following activity to solve the simultaneous equations
Image
→
Write the correct number in the given boxes from the following A. P.
– 3, – 8, – 13, – 18, . . .
$\text { Here } t _3=\square, t _2=\square, t _4=\square, t _1=\square $
$ t _2- t _1=\square, t _3- t _2=\square $
$ \therefore a =\square, d =\square$
→
The coordinates of the vertices of a triangle $ABC$ are $A (-7,6), B (2,-2)$ and $C (8,5)$. Find coordinates of its centroid.
Solution: Suppose $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$
$x_1=-7, y_1=6 \text { and } x_2=2, y_2=-2 \text { and } x_3=8, y_3=5$
Using Centroid formula
$\therefore$ Coordinates of the centroid of a traingle
$ ABC =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$
$=\left(\frac{\square}{3}, \frac{\square}{3}\right) $
$\therefore$ Coordinates of the centroid of a triangle $A B C=\left(\frac{3}{3}, \square\right)$
$\therefore$ Coordinates of the centroid of a triangle $A B C=(1, \square)$
→