Progression (Sequence and Series)Hard
Question
If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is
Options
A.
B.
C.
D.
Solution
tn = c {n2 - (n - 1)2}
= c (2n - 1)
⇒ tn2 = c2 (4n2 - 4n + 1)
⇒
=
{4(n + 1)(2n + 1)- 12(n + 1) + 6}
=
{4n2 + 6n + 2 - 6n - 6 + 3} = 
n(4n2 - 1)
= c (2n - 1)
⇒ tn2 = c2 (4n2 - 4n + 1)
⇒
=
{4(n + 1)(2n + 1)- 12(n + 1) + 6}=
{4n2 + 6n + 2 - 6n - 6 + 3} = n(4n2 - 1)Create a free account to view solution
View Solution FreeMore Progression (Sequence and Series) Questions
The geometric mean of the first n terms of the G.P. a + ar + ar2 +... is-...Product of n positive numbers is unity.The sum of these numbers can not be less than-...If the numbers a,b,c,d,e form an A.P., then the value of a − 4b + 6c − 4d + e is-...The sum of an infinitely decreasing G.P. is equal to 4 and the sum of the cubes of its terms is equal to 64 / 7. Then 5t...If in the arithmetic progression $\left\{ a_{n} \right\},a_{1} > 0,3a_{8} = 5a_{13}$, then among the following listed...