Question

i) If for a sequence $(t_{n}), S_{n}=2n^{2}+5n$, find $(t_{n})$ and show that the sequence is an A.P.
ii) If for a sequence $(t_{n}), S_{n}=4n^{2}-3n$, show that the sequence is an A.P.

Answer 1

Solution:

1) To find $(t_{n})$,put the values of n = 1,2,3,4...

$S_{n}=2n^{2}+5n \\ \\ for \: n = 1 \\ \\ S_{1}=2(1)^{2}+5(1) \\ \\ S_{1} = 7 \\ \\ for \: n = 2 \\ \\ S_{2}=2(2)^{2}+5(2) \\ \\ S_{2} = 8 + 10 \\ \\ S_{2} = 18 \\ \\ for \: n = 3 \\ \\ S_{3}=2(3)^{2}+5(3) \\ \\ S_{3} = 18 + 15 \\ \\ S_{3} = 33$
Here
$S_{1} = a = 7 \\ \\ S_{2} = a + t2= 18 \\ \\ t2 = 18 - 7 \\ \\ t2 = 11 \\ \\ S_{3} = a + t2 + t3 = 33 \\ \\ t3 = 33 - 18 \\ \\ t3 = 15$
So, these terms are 7,11,15...

common difference is d= 4
first term a = 7

Hence given expression is sum of AP,and these terms are AP
$t_{n} = a + (n - 1)d \\ \\ t_{n} = 7 + (n - 1)4 \\ \\ = 7 + 4n - 4 \\ \\t_{n}= 3 + 4n$

ii) If for a sequence $(t_{n}), S_{n}=4n^{2}-3n$, show that the sequence is an A.P.

Follow all the steps state above
$S_{n}=4n^{2}-3n \\ \\ for \: n = 1 \\ \\ S_{1}=4(1)^{2}-3(1) \\ \\ S_{1} = 1 = a \\ \\ for \: n = 2 \\ \\ S_{2}=4(2)^{2}-3(2) \\ \\ S_{2}=16-6 \\ \\ S_{2} = 10 = a + t_{2} \\ \\ 10 = 1 + t_{2} \\ \\ t_{2} = 9 \\ \\ for \: n = 3 \\ \\ S_{3}=4(3)^{2}-3(3) = 36 - 9 = 27 \\ \\ a + t_{2} + t_{3} = 27 \\ \\ 10 + t_{3} = 27 \\ \\ t_{3} = 17$

terms are 1,9,17...

a= 1

Common difference d = 8

Hence given expression of sum is created an AP,So

$t_{n} = a + (n - 1)d \\ \\ = 1 + (n - 1)8 \\ \\ = 1 + 8n - 8 \\ \\ t_{n}= - 7 + 8n$
Hope it helps you.