Question

The sum of first 10 terms of an ap is -150 and the sum of its next 10 terms is -550 find the AP.

Answer 1

Given,
S10=-150=T1+T2+T3+...........+T10
S20- S10=-550=T11+T12+T13+..........+T20
To find,
The AP
ATQ,
S10=10/2[2a+(10-1)d]
-150=5[2a+9d]
-30=2a+9d .......(i)
Again,
T1+T2+T3+......+T20=S20
S20=-150+(-550)
=-150-550
=-700
Now,
S20=20/2[2a+(20-1)d]
-700=10[2a+19d]
-70=2a+19d ........(ii)
Now,
Subtracting eq. (i) from eq. (ii):-
2a+19d-(2a+9d)=-70-(-30)
2a+19d-2a-9d=-70+30
10d=-40
d=-4
Now,
2a+9d=-30
2a+9(-4)=-30
2a-36=-30
2a=-30+36
2a=6
a=3
Then,
T1=a=3
T2=a+d=3+(-4)=-1
T3=a+2d=3+2(-4)=-5
Then,
AP:-3, -1, -5...........

Hope this helps you!

Answer 2

an = a1 + (n - 1)d

Sn = n/2 (2a + (n - 1)d)

Find the sum of the first 20 terms:

Sum of first 10 terms = -150

Sum of the next 10 term = -550

Sum of the first 20 term = -150 + (-550) = -700

Sum of first 10 terms is -150:

-150 = 10/2 (2a + 9d)

-150 = 5(2a + 9d)

-30 = 2a + 9d ------------------- [ 1 ]

Sum of first 20 terms is -700:

-700 = 20/2 (2a + 19d)

-700 = 10(2a + 19d)

-70 = 2a + 19d ------------------- [ 2 ]

Put the two equations together:

-30 = 2a + 9d ------------------- [ 1 ]

-70 = 2a + 19d ------------------ [ 2 ]

[ 2 ] - [ 1 ] :

-70 - {-30) = 19d - 9d

-40 = 10d

d = -40 ÷ 10

d = -4

Find a:

From [ 1 ]

-30 = 2a + 9(-4)

-30 = 2a - 36

2a = 6

a = 3

Form the AP:

a = 3

d = -4

an = a1 + (n - 1)d

an = 3 + (n - 1)(-4)

an = 3 - 4n + 4

an = 7 - 4n

Find the AP series:

a1 = 3

a2 = 7 - 4(2) = -1

a3 = 7 - 4(3) = -5

The AP series is 3, -1, -5, ....

Answer: The AP is 7 - 4n