Question

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

Answer 1

GIVEN :

11th term of an AP = 38

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

16th term of an AP = 73

a + 15d = 73 ------(2)

Solve equations 1 & 2 to find the first term the A.P (a)

a + 10d = 38

a + 15d = 73

(-)

-------------------

-5d = -35

d = 35/5

d = 7

Difference = 7

Substitute d in eq - 1

a + 10d = 38

a + 10(7) = 38

a + 70 = 38

a = 38 - 70

a = -32

31st term :

= a + 30d

= (-32) + 30(7)

= -32 + 210

= 178

Therefore, 178 is the 31st term of the AP.

Answer 2

Answer :-

Given :-

$\implies{\sf{{a}_{11}= 38}}$

$\implies{\sf{{a}_{16} = 73}}$

To Find :-

$\sf{The\ 31st\ term\ of\ A.P.}$

Solution :-

$\implies{\sf{{a}_{11} = 38}}$

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

$\implies{\sf{a + 10d = 38 .....(1)}}$

$\implies{\sf{and, {a}_{16} = 73}}$

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

$\implies{\sf{a + 15d = 73 .....(2)}}$

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

$\sf{a + 15d = 73}$

$\sf{a + 10d = 38}$

$\sf{[Multiply\ the\ (1)\ equation\ by (-)]}$

$\sf{ a + 15d = 73}$

$\sf{-a - 10d = -38}$

______________

$\sf{5d = 35}$

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

$\implies{\sf{d = 7}}$

Putting the value of d in (1) :-

$\implies{\sf{a + 10 (7) = 38}}$

$\implies{\sf{a = 38 - 70}}$

$\implies{\sf{a = -32}}$

Hence,

$\sf{{a}_{31} = a + (31 - 1) \times 7}$

$\implies{\sf{{a}_{31} = -32 + 210 = 178}}$

So,

$\boxed{\sf{The\ 31st\ term\ of\ A.P.\ is\ 178.}}$