Question

find the 11th term of the ap whose first two terms are -3,4​

Answer 1

Answer:

a11=a+(n-1)d

a11=-3+(11-1)7

a11=-3+70

a11=67........

Answer 2

The sum of first 20 terms of an A.P is -270.

Step-by-step explanation:

To find : The sum of first 20 terms of an AP in which 11th term is 5 and 13th term is 79 ?

Solution :

The nth term of an A.P is

$an \: = a + (n - 1)d$

The 11th term is 5.

$ie \: a + 10d = 5$

The 13th term is 79.

$ie \: a + 12d = 79$

$Subtract (1) from (2),$

$a + 12d - a - 10d = 79 - 5$

$2d = 74$

$d = \frac{74}{2}$

$d = 37$

$Substitute \: in (1),$

$a + 10(37) = 5$

$a = 5 - 370$

The sum of n terms of an A.P is

$sn = \frac{n}{2} |2a + (n - 1) d \: |$

The sum of first 20 terms of an A.P is

$s20 = - 270$

The sum of first 20 terms of an A.P is -270.