Question

The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?​

Answer 1

$\Large\mathfrak{\underline{\underline{Answer:-}}}$

➛ 150

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

$\small\mathfrak{\underline{\underline{Step-By-Step \:Explanation -}}}$

Given-

Sum of 4th and 12th term = 20

To find-

Sum of 1st 15 term of the AP

We know

$\Large\bold{\boxed{N'th\:term\:of\:AP= a + (n-1)d}}$

$\bf\Large\boxed{Sum\:of\:\:AP= \frac{n}{2}[2a+(n-1)d]}$

A/q to the question

Sum of 4th and 12th term = a +(4-1)d + a + 11d = 20

⇒ 2a +14d = 20 ______(1)

Sum of 1st 15 terms

$\bf\Large\boxed{\therefore\:S_{15}= \frac{15}{2}[2a+(15-1)d]}$

⇒ 15/2 (2a + 14d)

⇒ 15/2 × 20 [ from eqn 1]

⇒ 150

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

➛Arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant.

➛For Example, the sequence 3, 5, 7, 9, 11,.. is an arithmetic progression with common difference 2.

Answer 2

150

Given:

Value of n to find the sum of 15 terms = 15

The sum of the fourth and twelfth term of the AP = 20

The sum of the fourth and twelfth term of the AP is:

a + 3d  + a + 11d = 20

2 a + 14 d = 20

Formula used to find the sum of n terms of an arithmetic progression:

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

Now substituting the values which we know into this formula we get:

= 15/2(2a + 14d)

= 15/2 (20)

= 15 x 10

= 150

Therefore, the sum of 15 terms of the AP is 150.