Question

In an AP sum of 1st 10 term is -150 & the sum of the next 10 term is -550.Find AP.

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Can anybody answer...

Answer 1

In an AP sum of first 10 term is - 150.

Here..

• Number of terms (n) = 10

• $S_{10}$ = - 150

____________ [ GIVEN ]

Now..

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

Put the known values in above formula

=> $S_{10}$ = $\dfrac{10}{2}$ [2a + (10 - 1)d]

=> - 150 = 5 (2a + 9d)

=> - 30 = 2a + 9d

=> 2a + 9d = - 30 _________ (eq 1)

______________________________

Sum of next 10 term is - 550.

Here..

• First term (a) = a + 10

• Number of term (n) = 10

• $S_{10}$ = - 550

____________ [ GIVEN ]

Now..

=> $S_{10}$ = $\dfrac{10}{2}$ [2(a + 10d) + (10 - 1)d]

=> - 550 = 5 (2a + 20d + 9d)

=> - 110 = 2a + 29d

=> 2a + 29d = - 110 __________ (eq 2)

_______________________________

Subtract (eq 2) from (eq 1)

=> 2a + 29d - (2a + 9d) = - 110 - (- 30)

=> 2a + 29d - 2a - 9d = - 110 + 30

=> 20d = - 80

=> d = - 4

Put value of d in (eq 1)

=> 2a + 9(-4) = - 30

=> 2a - 36 = - 30

=> 2a = 6

=> a = 3

_______________________________

AP :

a = 3

a + d = 3 - 4 = - 1

a + 2d = 3 + 2(-4) = 3 - 8 = - 5

a + 3d = 3 + 3(-4) = 3 - 12 = - 9

______________________________

AP : 3, -1, -5, -9 ....

_____________ [ ANSWER ]

______________________________

Answer 2

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION:}}}$

Let a & d be the first & common difference of A.P.

S10= -150

$= > \frac{10}{2} (2a + (10 - 1)d) = - 150 \\ = > 2a + 9d = - 30........(1) \\ \\ a11 + a12 + a13 + .....a20 = - 550 \\ = > (a1 + a2 + ..... + a20) - (a1 + a2 + a3 + ..... + a20) = - 550 \\ = > a1 + a2 + .... + a20 - ( - 150) = - 550$

S20= -700

$= > \frac{20}{2} (2a + (20 - 1)d) = - 700 \\ = > 2a + 19d = - 70........(2)$

Solving equation (1) & (2), we get

(2a+19d)- (2a+9d)= (-70)-(-30)

=)10d= -40

=) d= -40/10

=) d= -4

When d= -4,we get

2a+9×(-4)= -30 [from (1)]

=) 2a-36= -30

=) 2a= -30+36

=) 2a= 6

=) a= 6/2

=) a= 3

Thus, the A.P. is 3, -1, -5,......

Hope it helps ☺️