Question

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

Answer 1

a +10d = 38
And
a + 15d = 73

Thus solving these two - subtracting one from two we get
15d - 10d = 73 - 38
5d = 35
Thus d = 7

Now substituting this in one of the equations we get
a + 10*7 =38
Thus a = 38 - 70 = -32

Thus 31st term of AP is
a + 30d = -32 + 30*7 = -32 + 210= 178

Answer 2

Given:

✫11th term of an A.P. = 38

✫16th term of an A.P. = 73

Find:

✫31st term will be = ?

Solution:

we, know that

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

For 11th term

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

$\sf \to 38 = a + (10)d$

$\sf \to a + 10d = 38......(i)$

For 16th term

$\sf \to a_{16} = a + (n - 1)d$

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

$\sf \to 73 = a + (15)d$

$\sf \to a + 15d = 73....(ii)$

Subtract eq(ii) from eq(i)

$\sf a + 10d = 38 \\ \underline{\sf - \boxed{ + }a - \boxed{ + }15d = - \boxed{ + }73} \\ \sf - 5d = - 35 \\ \sf d = \dfrac{ - 35}{ - 5} = 7 \\ \sf d = 7$

Put this value of d in eq(i)

$\sf \to a + 10d = 38$

$\sf \to a + 10(7) = 38$

$\sf \to a + 70= 38$

$\sf \to a = 38 - 70 = - 32$

$\sf \to a = - 32$

So,

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

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

$\sf \to a_{31} = - 32 + (30)(7)$

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

$\sf \to a_{31} = 178$

________________

Hence, the 31st term of an A.P. will be 178