Question

the A.P in which fourth term is -15 and 9th term is -30 find the sum of the first 10 numbers

Answer 1

Hii friend,

4th term = -15

a+3d = -15…....(1)

And,

9th term = -30

a +8d = -30......(2)

From equation (1) we get,

a + 3d = -15

a = -15-3d.....(3)

Putting the value of a in equation (2)

a + 8d = -30

-15-3d +8d = -30

5d = -30+15

5d =-15

D = -15/5 =-3

Putting the value of D in equation (3)

a = -15-3d => -15 - 3 × -3 = -15 +9 = -6

First term = -6

Second term = a+d = -6+(-3) = -6-3 = -9

Third term = a+2d = -6 + 2 × -3 = -12.

Fourth term = -15

fifth term = -18

Sixth term = -21

Seventh term = -24

Eighth term = -27

Nineth term = -30

And,

Tenth term = -33

Therefore,

The sum of first 10th term = -6-9-12-15-18-21-24-27-30-33 = -195

Hence,

The Sum of first 10th term of an AP = -195

HOPE IT WILL HELP YOU..... :-)

Answer 2

Fourth term = -15

a + 3d = -15

a = -15 -3d -----1equation

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9th term = -30

=> a + 8d = -30

Putting the value of a from 1equation

=> -15 -3d + 8d = -30

=> 5d = -30+15

=> 5d= -15

=> d = -15/5

=> d = -3

------------------------

a = -15 - 3(3)

a = -15 -3(-3)

a = -15 +9

a = -6

================

10th term = a + 9d =
-6+9(-3)=-6-27=-33

10th term = -33

___________________

Sum of n terms = n/2 [(first term + last term)

Sum of 10 term = 10/2 [-6+(-33)]

Sum of 10 terms = 5×-39

Sum of 10 terms = -195

I hope this will help you

-by ABHAY