Question

Find the sum of first 15th terms of the following APs : 11, 6, 1, -4, -9 .........​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore s_{15}=-285}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{given : } \\ \implies A.P = 11,6,1,- 4, - 9,..... \\ \\ \implies First \: term(a) = 11 \\ \\ \implies Common \: difference(d) = 6 - 11 = - 5 \\ \\ \underline \bold{To \: Find : } \\ \implies s_{15} = ?$

• According to given question :

$\bold{Using \: formula \: sum \: of \: n_{th} \: terms : } \\ \implies s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ \bold{Putting \: given \: values : }\\ \implies s_{15} = \frac{15}{2} (2 \times 11 + (15 - 1) \times -5 \\ \\ \implies s_{15} = \frac{15}{2} (22 + 14 \times - 5) \\ \\ \implies s_{15} = \frac{15}{2} (22 - 60) \\ \\ \implies s_{15} = \frac{15}{ \cancel2} \times \cancel{ - 38} \\ \\ \implies s_{15} = 15 \times - 19 \\ \\ \bold{\implies s_{15} = - 285}$

Answer 2

$\boxed{\mathsf{\green{Answer}}}$

Sum of 15 terms = -360

$\boxed{\mathsf{\green{Explanation}}}$

Given:

A.P. series is,

11, 6, 1, -4, -9...

To Find:

Sum of 15 terms of A.P.

Solution:

The A.P. is

11, 6, 1, -4...

Here,

a = 11

d = 6-11 = -5

n = 15

w.k.t

$\boxed{\bold{\pink{S_n = \dfrac{n}{2}[2a+(n-1)d]}}}$

$\mathsf{S_{15} = \dfrac{15}{2}[2(11)+(15-1)(-5)]}$

$\mathsf{S_{15} = \dfrac{15}{2}[22+(14)(-5)]}$

$\mathsf{S_{15} = \dfrac{15}{2}[22-70]}$

$\mathsf{S_{15} = \dfrac{15}{2}[-48]}$

$\mathsf{S_{15} = 15 (-24)}$

$\mathsf{S_{15} = -360}$