Question

the sum of first 6 term of an ap is 42.the ratio of its 11th term of its 33th term is 1:3.calculate the first term of ap

Answer 1

first term is 2
hope it helped u

Answer 2

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

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

$i) Sum \:of \: first \:6 \:terms = 42 \: (given)$

$\implies \frac{6}{2}[2a+(6-1)d] = 42$

$\implies 3(2a+5d) = 42$

$\implies 2a+5d = \frac{42}{3}$

$\implies 2a+5d = 14\: --(1)$

$\boxed {\orange {n^{th} \:term (a_{n}) = a + (n-1)d }}$

$Ratio \:of \: 11^{th} \:term \:of \:it's \\33^{th} \:term \:is \: 1 : 3\: (given)$

$\implies \frac{11^{th} \:term}{33^{th} \:term} = \frac{1}{3}$

$\implies \frac{a+10d}{a+32d} = \frac{1}{3}$

$\implies 3(a+10d) = a + 32d$

$\implies 3a+30d= a + 32d$

$\implies 3a - a = 32d - 30d$

$\implies 2a = 2d$

/* Dividing both sides by 2 ,we get */

$\implies a = d \: ---(2)$

$Now, 2a + 5d = 14 \: [Equation \:(1) ]$

$\implies 2a + 5a = 14 \: [ From \:(2) ]$

$\implies 7a = 14$

/* Dividing both sides by 7, we get */

$\implies a = 2$

Therefore.,

$\red { First \:term \:of \: given \:A. P }\green {= 2}$

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