Question

The A.P in which 4th term is -15 and 9th term is -30 find the sum of first 10 numbers

Answer 1

a4= (a+3d)= -15 ..... 1
a9= (a+8d)= -30...... 2

subtract 1 from 2

a+8d=-30
-a+3d=-15
-------------
0+5d=-15
d=-15/5
d=-3

put in 1
a+3d=-15
a-9=-15
a=-6

now, Sum of first 10 terms will be:
Sn=n/2(2a+(n-1)d)
S10=10/2(2(-6)+(10-1)-3)
S10=5(-12-27)
S10=5(-39)
S10=-195

So, the sum of first 10 terms of this A.P. is -195, hope it helps :)

Answer 2

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



D = -15/5 = -3


Putting the value of D in equation (3)


A = -15-3D = -15 - 3 × -3 = -15+9



A = -6





Therefore,





Sum of first 10th term =





Sn = N/2 × [ 2A + ( N -1) × D ]




S10 = 10/2 × [ 2 × -6 ( 10-1) × -3 ]






=> 5 ( -12 - 27)




=> 5 × -39




=> -195





★ HOPE IT WILL HELP YOU ★