Question

8. If a=10 and d=3 for an A.P. find the
*
sum of the first 15 terms.
Your answer​

Answer 1

Answer:

$\huge \boxed{s_{15} = 465}$

Step-by-step explanation:

→ Given:-

• a=10 (first term)
• d=3 (common difference)

→ To Find:-

$s_{15} \: (sum \: of \: 15 \: terms)$

→ Formula Used:-

$s_{n} = \frac{n}{2} \times [2a + (n - 1) \times d]$

→ Solution:-

$s_{n} = \frac{n}{2} \times [2a + (n - 1) \times d] \: ,where \: \boxed{a = 10}, \: \boxed{d = 3} \: and \: \boxed{ n = 15}$

$\implies s_{15} = \frac{15}{2} [2(10) + (15 - 1)(3)]$

$\implies s_{15} = \frac{15}{2} [2(10) + (14)(3)]$

$\implies s_{15} = \frac{15}{2} \times 2 [10 + (7)(3)]$

$\implies s_{15} = 15 (10 + 21)$

$\implies s_{15} = 15 \times 31$

$\implies \boxed{s_{15} = 465}$

Therefore,

Therefore,The Sum of 15 terms of this A.P. is 465.

→ One More formula to Remember:-

$a_n = a + ( n- 1) \times d$

Answer 2

Answer:

Step-by-step explanation:

Given

A= 10,d=3,S15=??

So

Formula = Sn= n/2(2a+(n-1)d)

By substituting we get

15/2(2×10+(15-1)3)

On solving .....

15/2(20+(14×3))

= 15/2(20+42)

=15×62/2

Ans= 465