Question

The aum of first n terma of an ap, whose first term is 6 andcommon difference is 40 equal to sum of two 2n terms of an ap whose first term is 40 and common difference is 6 find n

Answer 1

Answer:

11

Step-by-step explanation:

First AP

First term = 6

Common difference = 40

nth Term = First Term + (n-1) × Common Diferrence

=> nth Term = 6 + (n-1) × 40

=> nth term = 6 + 40n - 40

=> nth term = 40n -34

Sum of n term = (n/2) ( First Term + Nth Term)

=> Sum of n terms = (n/2)( 6 + 40n - 34)

=> Sum of n term = (n/2)(40n - 28)

=> Sum of  n terms = n(20n - 14)

=> Sum of n Terms = 20n² - 14n

Second  AP

First term = 40

Common difference = 6

2nth Term = First Term + (2n-1) × Common Diferrence

=> 2nth Term = 40 + (2n-1) × 6

=> 2nth term = 40 + 12n - 6

=> 2nth term = 12n + 34

Sum of 2n term = (2n/2) ( First Term + Nth Term)

=> Sum of 2n terms = (n)( 40 + 12n + 34)

=> Sum of 2n term = (n)(12n + 74)

=> Sum of  2n terms = 12n² + 74n

Hence

20n² - 14n = 12n² + 74n

=> 8n² = 88n  

=> 8n = 88   ( cancelling n both sides as n can not be zero)

=> n = 88/8

=> n = 11