# The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is – 30 and the common - Mathematics

Sum

The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is – 30 and the common difference is 8. Find n.

#### Solution

Given that, first term of the first AP (a) = 8

And common difference of the first AP(d) = 20

Let the number of terms in first AP be n

∵ Sum of first n terms of an AP

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

∴ S_n = n/2[2 xx 8 + (n - 1)20]

⇒ S_n = n/2 (16 + 20n - 20)

⇒ S_n = n/2(20n - 4)

∴ S_n = n(10n - 2)  .....(i)

Now, first term of the second AP(a’) = – 30

And common difference of the second AP(d’) = 8

∴ Sum of first 2n terms of second AP

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

⇒ S2n = n[2(– 30) + (2n – 1)(8)]

⇒ S2n = n[– 60 + 16n – 8)]

⇒ S2n = n[16n – 68] ………… (ii)

Now, by given condition,

Sum of first n terms of the first AP = Sum of first 2n terms of the second AP

⇒ Sn — S2n

⇒ n(10n – 2) = n(16n – 68)

From equation (i) and (ii)

⇒ n[(16n – 68) – (10n – 2)] = 0

⇒ n(16n – 68 – 10n + 2) = 0

⇒ n(6n – 66) = 0

⇒ n = 11  .......[∵ n ≠ 0]

Hence, the required value of n is 11.

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 33 | Page 54
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