Question

The 7th term of an ap 32 and its 13 th term is 62.Find the Ap.

Answer 1

⏺️Heya buddy ⏺

️$a7 = 32$
We know that,
An = a + (n-1)d

So,this can also be written as,
$a + (7 - 1)d = 32 \\ a + 6d = 32 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ........(i)$
Similarly,
$a13 = 62 \\ a + (13 - 1)d = 62 \\ a + 12d = 62 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .........(ii)$
Subtracting (i) from (ii),
$a + 12d - a - 6d = 62 - 32 \\ 6d = 30 \\ d = \frac{30}{6} \\ d = 5$
Substituting the value of 'd' in (i),
$a + 6(5) = 32 \\ a + 30 = 32 \\ a = 32 - 30 \\ a = 2$
.°. The AP will be
a, a + d, a + 2d, a + 3d....
=>2, 7, 12, 17....

♥️${Be } ^{Brainly }$♥️

Answer 2

The AP is 2,7, 12,17, ......