Question

Find the 31th term of an AP whose 11 term is 38 and the 16 term is 72.

Answer 1

Step-by-step explanation:

t11 =38 , t16 = 72

to find :t31

solution: tn=a +(n-1)d

t11=a+(11-1)d

t11=a+10d

38=a+10d (given)

a+10d=38 (1)

t16=a+(16-1)d

t16=a+15d

72=a+15d (given)

a+15d=72 (2)

subtracting 1 from 2

a+15d=72

a+10d=38

- - -

5d=34

d=34/5

substituting d=34/5 in (1)

a+10×34/5=38

a+68=38

a=30

t31= 30+(31-1)34/5

t3q=30+30×34/5

t31 =30+6×34

t31=30+ 204

t31=234

Answer 2

SOLUTION:-

Given:

11th term is 38.

16th term is 72.

We know that,

${}^{a} n = a + (n - 1)d$

Given 11th term is 38,

${}^{a} 11 = a + ( 11 - 1)d \\ \\ = > 38 = a + 10d \\ \\ = > a = 38 - 10d..............(1)$

Given 16th term is 72,

${}^{a} n = a + (16 - 1)d \\ \\ = > {}^{a} 16 = a + 15d \\ \\ = > 72 = a + 15d \\ \\ = > a = 72 - 15d..............(2)$

From equation (1) & (2), we get;

=) 38 - 10d = 72 -15d

=) 38 - 72= -15d +10d

=) -34 = -5d

=) d= 34/5

So,

Putting the value of d in equation (1), we get;

=) a= 38 -10d

$a = 38 - 10( \frac{34}{5} ) \\ \\ = > a = 38 - \frac{340}{5} \\ \\ = > a = 38 - 68 \\ \\ = > a = - 30$

Now,

We need to find 31th term.

⚫n= 31

⚫a= -30

⚫d= 34/5

${}^{a} n = a + (n - 1)d \\ \\ = > {}^{a} 31 = - 30 + (31 - 1)( \frac{34}{5} ) \\ \\ = > - 30 + (30)( \frac{34}{5} ) \\ \\ = > - 30 + 6 \times 34 \\ \\ = > - 30 + 204 \\ \\ = > 174$

Hence,

The 31th term of A.P. is 174.

Hope it helps ☺️