Question

1. 7th term of an AP is 40. The sum of its first 13th terms is
(a) 500 (b) 510 (c) 520 (d) 530​

Answer 1

7th term of an AP is 40. The sum of its first 13 terms is ....

(a) 500 (b) 510 (c) 520 (d) 530

solution : let first term is a and common difference is d.

nth term is given by, Tn = a + (n - 1)d

so, 7th term = a + (7 - 1)d = a + 6d = 40

⇒a + 6d = 40 ........(1)

now the sum of n terms , Sn = n/2 [2a + (n - 1)d]

so the sum of first 13 terms = S₁₃ = 13/2 [2a + (13 - 1)d ]

= 13/2 [2a + 12d ]

= 13 (a + 6d)

from equation (1) we get,

= 13 × 40

= 520

Therefore the sum of first 13 terms is 520 i.e., option (c) is correct choice.

Answer 2

Given ,

The 7th term of an AP is 40

We know that , the general formula of an AP is given by

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

Thus ,

40 = a + (7 - 1)d

40 = a + 6d

Now , the sum of first n terms of an AP is given by

$\boxed{ \sf S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}$

Thus ,

$\sf \mapsto S_{13} = \frac{13}{2} \{2a +12d \} \\ \\ \sf \mapsto S_{13} = \frac{13}{2} \times 2(a + 6d) \\ \\ \sf \mapsto S_{13} = 13 \times 40 \: \: \: \{ \because a + 6d = 40\} \\ \\ \sf \mapsto S_{13} =520$

Therefore ,

The sum of first 13 terms of given AP is 520