Question

The 7th term of an A.P. is 20 and its 13th term is 32. Find the A.P.​

Answer 1

Answer

A. P(arithmetic progression)

A. P(arithmetic progression) is 8,10 12,14,16.............

Given

in an Ap (arithmetic progression)

t7 = 20

t13=32

To find

A. P (arithmetic progression)

Explanation

we have given,

t7 = 20

we know, the formula of tn for finding the terms in an A.P

formula is,

tn = a + (n - 1) d

a stands for 1st term

d is for common difference

so,

t7 = a + ( n - 1)d

20 = a + ( 7 -1)d

20=a+6d ...........(1)

[consider it as a equation (1)]

Now, we have

t13=32

32= a + (13-1)d

32 = a+12d............(2)

[consider it as a equation (2)]

Substrate the equation (1) from (2)

32 = a + 12d

20 = a + 6d

_________________

12 = 6d

$\boxed{\textbf{\large{d=2}}}$

put [d=2] in eqn (1) we get,

20 = a + 6 (2)

20-12= a

therefor ,the value of a is

$\boxed{\textbf{\large{a=8}}}$

so, AP will be,

t1= a = 8

t2 = a + d = t1 + d = 8 + 2=10

t3 = t2 + d = 10 + 2 = 12

t4 = t3 + d = 12 + 2 = 14

t5 = t4 + d = 14 + 2 = 16

t6 = t5 + d = 16 + 2 = 18

t7 = t6 + d = 18 + 2 = 20

t1=8,t2=10,t3=12,t4=14......

so the AP (arithmetic progression) is = 8,10 12,14,16.............

Answer 2

Solution:

$7 {}^{th}term = 20$

$13 {}^{th} term = 32$

$common \: difference (d) = \frac{term \: difference}{position \: difference}$

$= \large \frac{32 - 20}{13 - 7}$

$= \large \frac{12}{6} = 2$

Common Difference = 2

$7 {}^{th} term = 20$

First term = 20 - 7d

= 20 - (7 × 2)

=20 - 14 = 6

So, the sequence is 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32...........