Question of RS Aggarwal book based on Arithmetic Progression For ICSE & CBSE Board Students.

Here the AP is given you have to find the value on which the *n* and the *n*th terms of AP are equal.

Question Number 21 Of Exercise 11A of RS Aggarwal Solution

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# For what value of n, the nth terms of arithmetic progression 63, 65,67,…..and 3,10,17,….are equal?

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In 1st AP:

Given AP is 63,65,67…..

First-term a = 63.

Common difference d = 65 – 63 = 2.

We know that sum of n terms an = a + (n – 1) * d

= 63 + (n – 1) * 2

= 63 + 2n – 2

= 61 + 2n ——- (1)

In 2nd AP :

Given AP is 3,10,17.

First-term a = 3.

Common difference d = 10 – 3 = 7.

We know that sum of n terms an = a + (n-1) d

= 3 + (n – 1) * 7

= 3 + 7n – 7

= 7n – 4 ——- (2)

Given that nth term of these AP are equal.

From (1) and (2),

we get

61 + 2n = 7n – 4

61 + 4 = 7n – 2n

61 + 4 = 5n

65 = 5n

n = 13.

Therefore 13th term of these AP’s are equal