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Question: The sum of first n terms of an A.P. is \(3n + {n^2}\) then find second, third and 15th term. (A). ...

The sum of first n terms of an A.P. is 3n+n23n + {n^2} then find second, third and 15th term.
(A). 6
(B). 8
(C). 32
(D). 45

Explanation

Solution

Hint- Sum of the first two terms times minus sum of the .first term will give us the second term, this can be done by substituting n=2 and 1. Similarly for third term substitute n=3 and 2 that is sum of three terms minus sum of first two terms, and for 15th take n=15 and 14.

Complete step-by-step answer:

Given that: sum of n terms of an A.P. is 3n+n23n + {n^2}
As we know that sum of n terms can be said as sum of n-1 terms and nth term. We will use this property to find the terms.
Second term of A.P.
From the given formula:
Sum of 2 terms of the given A.P. is:
∵Sn=3n+n2 ⇒S2=3(2)+22 ⇒S2=6+4 ⇒S2=10........(1)  \because {S_n} = 3n + {n^2} \\\ \Rightarrow {S_2} = 3\left( 2 \right) + {2^2} \\\ \Rightarrow {S_2} = 6 + 4 \\\ \Rightarrow {S_2} = 10........(1) \\\
Sum of 1 term of the given A.P. is:
∵Sn=3n+n2 ⇒S1=3(1)+12 ⇒S1=3+1 ⇒S1=4.........(2)  \because {S_n} = 3n + {n^2} \\\ \Rightarrow {S_1} = 3\left( 1 \right) + {1^2} \\\ \Rightarrow {S_1} = 3 + 1 \\\ \Rightarrow {S_1} = 4.........(2) \\\
So second term is equation (2) - equation (1)
T2=S2−S1 T2=10−4 T2=6  {T_2} = {S_2} - {S_1} \\\ {T_2} = 10 - 4 \\\ {T_2} = 6 \\\
So, the second term is 6.
Third term of A.P.
From the given formula:
Sum of 3 terms of the given A.P. is:
∵Sn=3n+n2 ⇒S3=3(3)+32 ⇒S3=9+9 ⇒S3=18.........(3)  \because {S_n} = 3n + {n^2} \\\ \Rightarrow {S_3} = 3\left( 3 \right) + {3^2} \\\ \Rightarrow {S_3} = 9 + 9 \\\ \Rightarrow {S_3} = 18.........(3) \\\
Sum of 2 terms of the given A.P. is:
∵Sn=3n+n2 ⇒S2=3(2)+22 ⇒S2=6+4 ⇒S2=10........(4)  \because {S_n} = 3n + {n^2} \\\ \Rightarrow {S_2} = 3\left( 2 \right) + {2^2} \\\ \Rightarrow {S_2} = 6 + 4 \\\ \Rightarrow {S_2} = 10........(4) \\\
So third term is equation (4) - equation (3)
T3=S3−S2 T3=18−10 T3=8  {T_3} = {S_3} - {S_2} \\\ {T_3} = 18 - 10 \\\ {T_3} = 8 \\\
So, the third term is 8.
Fifteenth term of A.P.
From the given formula:
Sum of 15 terms of the given A.P. is:
∵Sn=3n+n2 ⇒S15=3(15)+152 ⇒S15=45+225 ⇒S15=270.........(5)  \because {S_n} = 3n + {n^2} \\\ \Rightarrow {S_{15}} = 3\left( {15} \right) + {15^2} \\\ \Rightarrow {S_{15}} = 45 + 225 \\\ \Rightarrow {S_{15}} = 270.........(5) \\\
Sum of 14 terms of the given A.P. is:
∵Sn=3n+n2 ⇒S14=3(14)+142 ⇒S14=42+196 ⇒S14=238........(6)  \because {S_n} = 3n + {n^2} \\\ \Rightarrow {S_{14}} = 3\left( {14} \right) + {14^2} \\\ \Rightarrow {S_{14}} = 42 + 196 \\\ \Rightarrow {S_{14}} = 238........(6) \\\
So fifteenth term is equation (5) - equation (6)
T15=S15−S14 T15=270−238 T15=32  {T_{15}} = {S_{15}} - {S_{14}} \\\ {T_{15}} = 270 - 238 \\\ {T_{15}} = 32 \\\
So, the fifteenth term is 32.
Hence, the second term of the A.P. is 6. Third term of the A.P. is 8. Fifteenth term of the given A.P. is 32.
So, options A, B, C are the correct options.

Note- Another approach-Substitute the value of n as 1 and 2 to get the sum of the first term and first two terms, subtract to find the second term, now from first and second term calculate common difference, now use the nthn^{th} term formula of an AP to find the required terms.