Q18 of 38 Page 5

In an AP, if then find the AP.

Given: Sn = n(4n+1)


To find: The AP


Formula used:


Sn - Sn – 1= an



Explanation:


Sn = n (4n + 1)


Sn - 1 = (n - 1) [ 4(n - 1) - 1]


= (n - 1) [ 4n - 5]


Sn - Sn - 1 = n (4n + 1 ) - ( n - 1 )( 4n - 5 )


⇒ (a1 + a2 + a3 + - - - + an - 1 + an ) - ( a1 + a2 + a3 + - - - + an - 1 )


= 4n2 + n - (4n2 - 5n - 4n + 5 )


an = 11n - 5


For a1,


Put n = 1 so a1 = 11(1) - 5 = 6


For a2,


Put n = 2, so a1 = 11(2 ) - 5 = 17


For a1,


Put n = 3 so a1 = 11(3 ) - 5 = 28


For a1,


Put n = 4 so a1 = 11(4 ) - 5 = 39


So, AP is 6, 17, 28, 39, …


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