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Sum of an AP in general with the first term = a and upto n terms:
Sn = n/2 * [2a +(n-1)d ] = n/2 * [2a - d + d n] ---(1)
Given sum upto n terms: 2n² + 3n = n/2 * [ 6 + 4 n ] ---(2)
Comparing (1) and (2) we get:
d = 4.
2 a - d = 6 => a = 5.
So 6th term = a + (m-1) d
= 5 + (6-1) 4 = 25
16th term = a + 15 * d = 5 + 15*4 = 65
Sn = n/2 * [2a +(n-1)d ] = n/2 * [2a - d + d n] ---(1)
Given sum upto n terms: 2n² + 3n = n/2 * [ 6 + 4 n ] ---(2)
Comparing (1) and (2) we get:
d = 4.
2 a - d = 6 => a = 5.
So 6th term = a + (m-1) d
= 5 + (6-1) 4 = 25
16th term = a + 15 * d = 5 + 15*4 = 65
ajj2:
I thing its wrong
which one you asked 16th term or 6th term ? check your answers
it is good and right
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