Question

find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73 ?

Answer 1

let first term of Ap is a and common difference is d

use formula

tn=a+(n-1) d

now

t11=a+(11-1) d=a+10d=38

same way

t16=a+15d=73

solve this equation then

a=-32 and d=7

now t31=a+30d=-32+210=178

Answer 2

Answer:

$\bold{\sf{\therefore a_{31} = 178}}$

Step-by-step explanation:

$\huge{\underline{\sf{SOLUTION-}}}$

$\implies a_{11} = 38 \\\\ \implies a + 10d = 38 - - - - - (1) \\ \\ \implies a_{16} = 73 \\ \\ \implies a + 15d = 73 - - - - - (2) \\ \\ \bold {subtracting \: (1) \: from \: (2) }\\ \\\implies a + 15d - (a + 10d) = 73 - 38 \\ \\ \implies a + 15d - a - 10d = 35 \\ \\ \implies 5d = 35 \\ \\ \implies d = \frac{35}{5} \\\\ { \bold{ \to d = 7}} \\ \\ \bold{ putting \: value \: of \: d \: in \: (1) }\\\\ \implies a + 10d = 38 \\ \\ \implies a + 10 \times 7 = 38 \\ \\ \implies a + 70 = 38 \\ \\ \implies a = 38 - 70 \\ { \bold{\implies a = - 32}}$

• We find the first term and common difference :

• So According to the question:

$\implies a_{31} = a + 30d \\\\ \implies a_{31} = - 32 + 30 \times 7 \\ \\ \implies a_{31} = - 32 \times 210 \\\\ { \bold{\therefore a_{31} = 178}}$