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In an AP given a = 8, an = 62, Sn = 210, find n and d. - Mathematics

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In an AP given a = 8, an = 62, Sn = 210, find n and 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.

n and d, if a = 8, an = 62 and Sn = 210

Sum

Solution 1

a = 8, an = 62 and Sn = 210 (Given)

∵ Sn = `"n"/2` [a + an]

⇒ 210 = `"n"/2 [8 + 62]`

⇒ 210 = `"n"/2` × 70

⇒ 35n = 210

⇒ n = `210/35`

⇒ n = 6

∵ an = a + (n - 1) × d

⇒ 62 = 8 + (6 - 1) × d

⇒ 62 = 8 + 5d

⇒ 5d = 62 - 8

⇒ 5d = 54

⇒ d = `54/5`

Hence, the required values of n and 4 are 6 and `54/5` respectively.

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Solution 2

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

Here,

First term (a) = 8

Last term (`a_n`) = 62

Sum of n terms (Sn) = 210

Now, here, the sum of the n terms is given by the formula,

`S_n = (n/2)(a + l)`

Where a is the first term

l = the last term

So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,

`210 = (n/2)[8 + 62]`

210(2) = n(70)

`n = 420/70`

n = 6

Also, here we will find the value of d using the formula,

an = a + (n - 1)d

So, substituting the values in the above mentioned formula

62 = 8 + (6 - 1)d

62 - 8 = (5)d

`54/5 = d`

`d = 54/5`

Therefore, for the given A.P `n = 6 and d = 54/5`.

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Chapter 5: Arithmetic Progressions - Exercise 5.3 [Page 112]

APPEARS IN

NCERT Mathematics [English] Class 10
Chapter 5 Arithmetic Progressions
Exercise 5.3 | Q 3.07 | Page 112
RD Sharma Mathematics [English] Class 10
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 56. 5

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