Question

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73

Answer 1

we know tat a 11 = 38 so , a+ 10d =38

a16 = 73 , a+15d = 73

by equating them by elimination method we get d =7

now substitute the vale of d in any of the eq.

we get a +10d=38

a +10[7] =38

a= -32

now we should find out the 31st term so

a31=a +[n-1]d

= -32 +30 * 7

=-32+210

=178

There forethe 31st term is 178...

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

$\textbf{\underline{First\;term\;be\;p}}$

$\textbf{\underline{Common\; Difference\;be\;d}}$

Using Formula

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

Here a = First term but we assume first term be p

Hence

p + (n - 1)d

$\tt{\rightarrow p_{11}=p+10d=39.....(1)}$

$\tt{\rightarrow p_{16}=p+15d=73.....(2)}$

${\boxed{\sf\:{Subtracting\;(1)\;from\;(2)}}}$

5d = 35

$\tt{\rightarrow d=\dfrac{35}{5}}$

d = 7

${\boxed{\sf\:{Put\;the\;value\;of\;d\;in\;(1)}}}$

p + 10(7) = 38

p + 70 = 38

p = 38 - 70

p = -32

Hence,

${\boxed{\sf\:{p_{31}=a+30d}}}$

= -32 + 30(7)

= -32 + 210

= 178

$\Large{\boxed{\sf\:{Hence\;31^{st}\;term\;is\;178}}}$