Question

given a = 7, a13 = 35, find d and Sn.​

Answer 1

GIVEN :

First term of an AP = a = 7

Thirteenth term of an AP = 35

a + 12d = 35 ------(1)

Substitute a in eq - (1)

a + 12d = 35

(7) + 12d = 35

12d = 35 - 7

12d = 28

d = 28/12

d = 7/3

Common Difference = 7/3

In an AP nth term = a + (n-1)d

35 = 7 + ( n - 1 )7/3

35 - 7 = (n - 1)7/3

28 = (n - 1)7/3

28 × 3 = 7n - 7

84 = 7n - 7

84 + 7 = 7n

91 = 7n

n = 91/7

n = 13

Number of terms = 13

In an AP sum of the terms = n/2 ( a + an )

= 13/2 ( 7 + 35)

= 13/2 ( 42)

= 13(21)

= 273

Therefore, the sum of the terms (sn) = 273.

Answer 2

a = 7, $a_{13}$ = 35

_________ [GIVEN]

• We have to find d and $S_{n}$

____________________________

First term (a) = 7

$a_{n}$ = a + (n - 1)d

=> $a_{13}$ = 7 + (13 - 1)d

=> 35 = 7 + 12d

=> 28 = 12d

=> d = $\dfrac{7}{3}$

_____________________________

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

OR

$S_{n}$ = $\dfrac{n}{2}$ (a + an)

Here ..

n = 13, a = 7, an = 35

=> $\dfrac{13}{2}$ (7 + 35)

=> $\dfrac{13}{2}$ × 42

=> 13 × 21

=> 273

____________________________

d = $\dfrac{7}{3}$ and $S_{n}$ = 273

_________ [ANSWER]