Question

In an



a.p sum of first 10 terms is _150 and the sum of its next term is _550 find tje



a.p

Answer 1

$Let \: a \:and \:d \: are \: first \:term \:and \\common \: difference \: of \:an \:A.P$

$a, (a+d), \cdot \cdot \cdot ,(a+9d) \:are \: first\\10 \:terms \: of \:A.P$

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

$i) Sum \:of \:10 \:terms (S_{10})\\=150 \:(given)$

$\implies \frac{10}{2}[2a+9d] = 150$

$\implies 5[2a+9d] = 150$

$\implies 2a+9d = \frac{150}{5}$

$\implies 2a+9d = 30 \: ---(1)$

$a+10d, a+11d, \cdot\cdot\cdot, a+19d \:are \\next \: 10 \:terms \: in \:A.P$

$ii) Sum \:of \:next\:10 \:terms (S_{10})\\=550 \:(given)$

$\implies \frac{10}{2}[2(a+10d)+9d] = 550$

$\implies 5[2a+20d+9d] = 550$

$\implies 2a+29d = \frac{550}{5}$

$\implies 2a+29d = 110 \: ---(2)$

/* Subtract equation (1) from equation (2) , we get */

$\implies 20d = 80$

$\implies d = \frac{80}{20}$

$\implies d = 4 \: --(3)$

/* Put d = 4 in equation (1) , we get */

$2a + 9 \times 4 = 30$

$\implies 2a + 36 = 30$

$\implies 2a = 30 - 36$

$\implies 2a = -6$

$\implies a = \frac{-6}{2}$

$\implies a = -3$

Therefore.,

$\red{ Required\:A.P }$

$\green { -3, 1, 5, 9, \cdot\cdot \cdot }$

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Answer 2

Refers to attachment ◇~