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In an AP given a = 3, n = 8, Sn = 192, find d.

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 In an AP given a = 3, n = 8, Sn = 192, find d.

Let there be an A.P. with the first term 'a', common difference 'd'. If an a denotes in nth term and Sn the sum of first n terms, find.

d, if a = 3, n = 8 and Sn = 192

Sum
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Solution 1

Given that, a = 3, n = 8, Sn = 192

`S_n = n/2 [2a+(n-1)d]`

`192 = 8/2[2xx3+(8-1)d]`

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

`d = 42/7`

d = 6

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Solution 2

Here, we have an A.P. whose first term (a), the sum of first n terms (Sn) and the number of terms (n) are given. We need to find the common difference (d).

Here,

First term (a) = 3

Sum of n terms (Sn) = 192

Number of terms (n) = 8

So here we will find the value of n using the formula, an = a + (a - 1)d

So, to find the common difference of this A.P., we use the following formula for the sum of n terms of an A.P

`S_n = n/2 [2a + (n -1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

`S_8 = 8/2 [2(3) + (8 - 1)(d)]`

192 = 4[6 + (7) (d)]

192 = 24 + 28d

28d = 192 - 24

Further solving for d

`d = 168/28`

d = 6

Therefore, the common difference of the given A.P. is d = 6.

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Chapter 5: Arithmetic Progressions - EXERCISE 5.3 [Page 68]

APPEARS IN

NCERT Mathematics [English] Class 10
Chapter 5 Arithmetic Progressions
EXERCISE 5.3 | Q 3. (ix) | Page 68
R.D. Sharma Mathematics [English] Class 10
Chapter 5 Arithmetic Progressions
Exercise 5.6 | Q 56. 3
ML Aggarwal Understanding Mathematics [English] Class 10 ICSE
Chapter 9 Arithmetic and Geometric Progressions
Exercise 9.3 | Q 4.5

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