Question

how many terms of an A.P 1,4,7,.....are needed to give the sum 1335​

Answer 1

Answer:

hope its help u......

Step-by-step explanation:

22 terms are needed to give sum 1335

s=n/2{2a+(n-1)d}

715=n/2{2X1+(n-1)3}

1430=n(2+3n-3)

3n^2-n-1430=0

3n^2-(66-65)n-1430=0

3n(n-22)-65(n-22)=0

(n-22)(3n-65)=0

=>n-22=0

n=22

Answer 2

$\textbf{\large{\pink{\underline{\underline{Step by Step Explanation:-}}}}}$

$\mathcal{\large{\underline{\ Given,,}}}$

$\mathtt{\small{\ a=1}}$

$\mathtt{\small{\ S_{n}=1335}}$

$\mathtt{\small{\ d = a_{2}-a}}$

$\mathtt{\small{\implies{\ d = 4-1}}}$

$\mathtt{\small{\implies{\ d=3}}}$

$\mathtt{\small{\ n=?}}$

$\textbf{Hence,,}$

$\mathtt{\small{\ S_{n}=\frac{n}{2}(2a+(n-1)d)}}$

$\mathtt{\small{\ 1335=\frac{n}{2}(2\times{1}+(n-1)3)}}$

$\mathtt{\implies{\small{\ 1335\times{2}=n(2+3n-3)}}}$

$\mathtt{\implies{\small{\ 2670={n(3n-1)}}}}$

$\mathtt{\implies{\small{\ 2670=3n^{2}-n}}}$

$\mathtt{\implies{\small{\ 3n^{2}-n-2670=0}}}$

$\mathtt{\implies{\small{\ 3n^{2}-90n+89n-2670=0}}}$

$\mathtt{\implies{\small{\ 3n(n-30)+89(n-30)=0}}}$

$\mathtt{\implies{\small{\ (n-30)(3n+89)=0}}}$

$\mathtt{\implies{\small{\ n-30=0\:or\:3n+89=0}}}$

$\mathtt{\implies{\small{\ n=30,,n=\frac{89}{3}}}}$

$\mathtt{n=\frac{89}{3}.. is..not..an..integer}$

$\mathbf{so,n=30}$

$\textbf{Hence,, There are 30 terms in AP..}$

$\textbf{\large{\green{Hope it helps you.. thanks}}}$

$\textbf{\large{\red{Please mark it as Brainliest answer.. thanks}}}$