Question

find the 31 term of an ap whose 11 term is 38 and 16 term is 73
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Answer 1

Given:-

• 11th term of an A.P = 38

• 16th term of an A.P = 73

To Find:-

• 31st term of an A.P

Formula Used:-

• ${\boxed{\bf{a_n=a+(n-1)d}}}$

Solution:-

As given,

$\sf :\implies\:a_{11}= 38$

$\sf :\implies\:a+(n-1)d= 38$

$\sf :\implies\:a+(11-1)d= 38$

$\sf :\implies\:a+10d= 38$

$\bf :\implies\:a= 38-10d$ ____{1}

And,

$\sf :\implies\:a_{16}= 73$

$\sf :\implies\:a+(n-1)d= 73$

$\sf :\implies\:a+(16-1)d= 73$

$\bf :\implies\:a+15d= 73$

Putting Value of a,

$\sf :\implies\:38-10d+15d= 73$

$\sf :\implies\:38+5d= 73$

$\sf :\implies\:5d= 73-38$

$\sf :\implies\:d=\dfrac{35}{5}$

$\bf :\implies\:d= 7$

Putting value of d in {1},

$\sf :\implies\:a= 38-10\times 7$

$\sf :\implies\:a= 38-70$

$\bf :\implies\:a= -32$

Now,

$\sf :\implies\:a_{31}=a+(n-1)d$

$\sf :\implies\:a_{31}=-32+(31-1)7$

$\sf :\implies\:a_{31}=-32+30\times7$

$\sf :\implies\:a_{31}=-32+210$

$\bf :\implies\:a_{31}=178$

Hence, 31st term of an A.P is 178.

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More Formulas related to A.P:-

• $\bf S_n=\dfrac{n}{2}[2a+(n-1)d]$

• $\bf S_n=\dfrac{n}{2}[a+a_n]$

• $\bf S_n=\dfrac{n}{2}[a+l]$

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