Question

if sum of first 11 terms of an ap is 33 then find the 6th term​

Answer 1

Answer:

This ur solution...I hope it helped..

Answer 2

$Let \: \pink {a }\: and \: \blue {d } \: are \: first \:term \: and \: Common \: differnce \:of \:an \: A.P$

$We \:know \: that ,$

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

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

$Here, Sum \:of \: first \:11 \:terms (S_{11}) = 33$

$\implies \frac{11}{2}[2a+(11-1)d] = 33$

$\implies \frac{11}{2}[2a+10d] = 33$

$\implies \frac{11\times 2}{2}[a+5d] = 33$

$\implies 11[ a + 5d ] = 33$

$\implies a + (6-1)d = \frac{33}{11}$

$\implies a_{6} = 3$

Therefore.,

$\red { 6^{th} \:term \: of \: A.P } \green {= 3 }$

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