In an AP, the sum of …

in continuation from above 

so AP is  3 , -1 , -5, -9, -13, -17 , -21, -25, -29, -33  

                             sum of 1st 10 terms = -150

-37 , -41, -45, -49, -53, - 57, -61, -65, -69, -73          

                               sum of next 10 terms = - 550

sorry due to power supply failure i could not complete

from equation i and ii above we get

2a + 9d = - 30

2a + 19d = - 70

- 10d = + 40

d = - 4

2a - 36 = - 30

2a = 6

a = 3

Let first term = a

common difference = d

According to question,

{tex}S_{10}= -150{/tex}

{tex}=> {10\over 2}[2a+9d]=-150{/tex}

{tex}=> 2a+9d= -30 \ ........(1){/tex}

Also

{tex}=> S_{20}-S_{10}=-550{/tex}

{tex}=> S_{20}+150=-550{/tex}

{tex}=> S_{20} = -700{/tex}

{tex}=> {20\over 2}[2a+19d]=-700{/tex}

{tex}2a+19d= -70\ .......(2){/tex}

Solving (1) and (2)

10 d = -40

=> d = -4

=> a = 3

{tex}So AP = 3,-1,-5,-9,-13,.. {/tex}

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

{tex}S_{10}= -150{/tex}

{tex}=> {10\over 2}[2a+9d]=-150{/tex}

{tex}=> 2a+9d= -30 \ ........(1){/tex}

Again

{tex}a_{11}+a_{12}+......+a_{20}= -550{/tex}

{tex}=> S_{20}-S_{10}=-550{/tex}

{tex}=> S_{20}+150=-550{/tex}

{tex}=> S_{20} = -700{/tex}

{tex}=> {20\over 2}[2a+19d]=-700{/tex}

{tex}2a+19d= -70\ .......(2){/tex}

Subtract (1) from (2), we get

10 d = -40

=> d = -4

Putting value of d in (1) we get

a = 3

So AP = 3,-1,-5,-9,....

s_{n = n/2 (2a + (n - 1)d)}

 _{-150 = 5(2a + 9d)      - 30 = 2a + 9d ....... i}

_{- 550 + -150 = 10(2a +19d)}

_{- 700 = 10 (2a + 19d.....ii}