Question

1.. If the nth term of an AP is (4n+1), find the sum of the first 15 terms of thi
AP. Also find the sum of its n terms.
2. The sum of the first n terms of an AP is given by Sn = (2n² +5n). Find the
nth term of the AP.
​

Answer 1

$\large{\underline{\tt{\purple{Given:-}}}}$

• ✦ nth term of AP is (4n+1)

$\large{\underline{\tt{\purple{To\:Find:-}}}}$

• ✦sum of 15th terms.

$\huge{\underline{\tt{\green{Solution:-}}}}$

• nth term of AP is (4n+1)

1st term = 4 ×1+1

⠀⠀⠀⠀⠀➝ 4 +1

⠀⠀⠀⠀⠀➝ 5

2nd term = 4 ×2 +1

⠀⠀⠀⠀⠀➝ 8 +1

⠀⠀⠀⠀⠀➝ 9

3rd term = 4×3 +1

⠀⠀⠀⠀⠀➝ 12 +1

⠀⠀⠀⠀⠀➝ 13

So,

AP = 5 ,9 ,13......

• First term (a) = 5
• common difference (d) = 9-5 = 4

Sum of 15th terms of AP :-

$\bf\green{S_n = \frac{n}{2}(2a + (n - 1)d)}$

$\mapsto \: \: \rm \: S_{15 }= \frac{15}{2}(2 \times 5 + (15- 1)4) \\ \\ \mapsto \: \: \rm \: S_{15 } = \frac{15}{2}(10+ 14 \times 4) \\ \\\mapsto \: \: \rm \: S_{15 } = \frac{15}{2} ( 10+ 56 ) \\ \\ \mapsto \: \: \rm \: S_{15 } = \frac{15}{{\cancel{2}}}\times{\cancel{66}} \\ \\ \mapsto \: \: \rm \: S_{15 } = 15 \times 33 \\ \\\mapsto \: \: \rm \: S_{15 } = 495 \\ \\\mapsto \: \: \bf \: \red{ S_{15 } = 495 }$

Sum of nth term:-

$\mapsto\rm\:S_n=\frac{n}{2}[2\times5+(n-1)4]\\\\ \mapsto \: \: \rm \: s_ n=\frac{n}{2}[10+4n-4]\\\\ \mapsto \: \: \rm \: s_ n=\frac{n}{2}[6+4n]\\\\ \mapsto \: \: \rm \: s_ n=3n+2n^2\\\\ \mapsto \: \: \bf \: s_ n=n(3+2n)$

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{1. \ The \ sum \ of \ first \ 15 \ terms \ of}$

$\sf{2. \ The \ n^{th} \ term \ of \ AP \ is \ (4n+3).}$

$\sf{AP \ is \ 495.}$

$\sf\orange{Given:}$

$\sf{\implies{1. \ n^{th} \ term \ of \ AP \ is \ (4n+1)}}$

$\sf{\implies{2. \ Sn=(2n^{2}+5n)}}$

$\sf\pink{To \ find:}$

$\sf{1. \ Sum \ of \ first \ 15 \ terms.}$

$\sf{2. \ n^{th} \ term \ of \ the \ A.P.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{1.}$

$\sf{\implies{tn=4n+1}}$

$\sf{\therefore{t1=4(1)+1}}$

$\sf{\implies{\therefore{t1=5}}}$

$\sf{t15=4(15)+1}$

$\sf{t15=60+1}$

$\sf{\implies{\therefore{t15=61}}}$

$\sf{Sn=\frac{n}{2}[t1+tn]... formula}$

$\sf{\therefore{S15=\frac{15}{2}[5+61]}}$

$\sf{\therefore{S15=\frac{15}{2}\times66}}$

$\sf{\therefore{S15=15\times33}}$

$\sf{\therefore{S15=495}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ first \ 15 \ terms \ of}}}$

$\sf\purple{\tt{AP \ is \ 495.}}$

$\sf{2.}$

$\sf{\implies{Sn=(2n^{2}+5n)...(1)}}$

$\sf{\therefore{S1=[2(1)^{2}+5(1)]}}$

$\sf{\therefore{S1=2+5}}$

$\sf{\implies{\therefore{S1=7}}}$

$\sf{\implies{\therefore{t1=7}}}$

$\sf{Sn=\frac{n}{2}[t1+tn]…formula...(2)}$

$\sf{From \ (1) \ and \ (2)}$

$\sf{\therefore{2n^{2}+5n=\frac{n}{2}[t1+tn]}}$

$\sf{\therefore{n(2n+5)=\frac{n}{2}[7+tn]}}$

$\sf{\therefore{\frac{2}{n}\times \ n(2n+5)=7+tn}}$

$\sf{\therefore{7+tn=2(2n+5)}}$

$\sf{\therefore{7+tn=4n+10}}$

$\sf{\therefore{tn=4n+10-7}}$

$\sf{\implies{\therefore{tn=4n+3}}}$

$\sf\purple{\tt{\therefore{The \ n^{th} \ term \ of \ AP \ is \ (4n+3).}}}$