Question

How many terms of the arithmetic series 6 + 13 + 20 + 27 +...... Be taken continuously so that their sum is 375?

Answer 1

Here is your answer ok

Answer 2

$Given \: A.P : 6,13,20,27, \cdot\cdot\cdot$

$First \:term (a) = 6$

$Common \: difference (d) = a_{2} - a_{1} \\= 13 - 6 \\= 7$

$6 + 13 + 20 + 27 + \cdot \cdot \cdot ,n \:terms = 375$

$\boxed { \pink { Sum \:of \: n \:terms (S_{n}) = \frac{n}{2}[2a+(n-1)d] }}$

$\implies \frac{n}{2}[2\times 6 +(n-1)7 ] = 375$

$\implies n(12 +7n-7)= 750$

$\implies n(7n+5)= 750$

$\implies 7n^{2}+5n - 750= 0$

/* Splitting the middle term, we get */

$\implies 7n^{2}+75n -70n - 750= 0$

$\implies n(7n+75) - 10(7n+75) = 0$

$\implies (7n+75)(n-10) = 0$

$\implies 7n+75= 0 \: Or \: n-10 = 0$

$\implies n = \frac{-75}{7} \: Or \: n = 10$

/* n should not be negative */

$\implies n = 10$

Therefore.,

$\green { Sum \: of \: 10 \: terms \: in }\\\green {given \:A.P \: is \: 375 }$

•••♪

•••♪