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Find the sum of first 17 terms of an AP whose 4th and 9th terms are -15 and -30 respectively

Solution:

Given, the 4-th term of an AP = -15

The 9-th term of an AP = -30

We have to find the first 17 terms of an AP.

The nth term of the series in AP is given by

aₙ = a + (n - 1)d

When n = 4, a₄ = a + (4 - 1) d

a + 3d = -15 --------------- (1)

When n = 9, a₉ = a + (9 - 1)d

a + 8d = -30 ----------------(2)

Subtracting (1) from (2),

a + 8d - (a + 3d) = -30 - (-15)

a + 8d - a - 3d = -30 + 15

8d - 3d = -15

5d = -15

d = -15/5

d = -3

Put d = -3 in (1),

a + 3(-3) = -15

a - 9 = -15

a = -15 + 9

a = -6

The sum of the first n terms of an AP is given by

Sₙ = n/2[2a + (n-1)d]

To find the sum of the 17 terms,

S₁₇ = 17/2[2(-6) + (17 - 1)(-3)]

= 17/2[-12 + (16)(-3)]

= 17/2[-12 - 48]

= 17/2[-60]

= 17(-30)

S₁₇ = -510

Therefore, the sum of the first 17 terms is -510.

✦ Try This: Find the sum of the first 17 terms of an AP whose 4th and 9th terms are 8 and 18

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 27

Find the sum of first 17 terms of an AP whose 4th and 9th terms are -15 and -30 respectively

Summary:

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same. The sum of the first 17 terms of an AP whose 4th and 9th terms are -15 and -30 is -510

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