If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms. - Mathematics

Sum

If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Solution

Let a and d be the first term and common difference, respectively of an AP.

∵ Sum of n terms of an AP

S_n = n/2 [2a + (n - 1)d]  .....(i)

Now, S_6 = 36  ......[Given]

⇒ 6/2[2a + (6 - 1)d] = 36

⇒ 2a + 5d = 12  ......(ii)

And S_16 = 256

⇒ 16/2[2a + (16 - 1)d] = 256

⇒ 2a + 15d = 32  .......(iii)

On subtracting equation (ii) from equation (iii), we get

10d = 20

⇒ d = 2

From equation (ii)

2a + 5(2) = 12

⇒ 2a = 12 − 10 = 2

⇒ a = 1

∴ S_10 = 10/2 [2a + (10 - 1)d]

= 5[2(1) + 9(2)]

= 5(2 + 18)

= 5 × 20

= 100

Hence, the required sum of first 10 terms is 100.

Concept: Sum of First n Terms of an A.P.
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APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.3 | Q 28 | Page 54
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