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The 4th term of an A.P. is 22 and the 15th term is 66. Find the first terns and the common difference. Hence find the sum of the series to 8 terms. - Mathematics

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The 4th term of an A.P. is 22, and the 15th term is 66. Find the first term and the common difference. Hence, find the sum of the series to 8 terms.

The 4th term of an A.P. is 22, and the 15th term is 66. Find the sum of its 8 terms.

Sum
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Solution

Let a be the first term and d be the common difference of the given A.P.

Now,

4th term = 22

⇒ a + 3d = 22 ...(i)

15th  term = 66

⇒ a + 14d

= 66

Subtracting (i) from (ii), we have

11d = 44

⇒ d = 4

Substituting the value of d in (1), we get

a = 22 − 3 × 4

= 22 − 12

=10

⇒ First term = 10

Now

Sum of 8 terms = `8/2[2xx10+7xx4]`

= 4[20 + 28]

= 4 × 48

= 192

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Chapter 9: Arithmetic and geometric progression - Exercise 9C [Page 187]

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Nootan Mathematics [English] Class 10 ICSE
Chapter 9 Arithmetic and geometric progression
Exercise 9C | Q 8. | Page 187
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