in an AP the sum of the first 10 terms is -150. And the sum of next 10 terms is -550. Find the AP.

in an AP the sum of the first 10 terms is -150. And the sum of next 10

1.	in an AP the sum of the first 10 terms is -150. And the sum of next 10 terms is -550. Find the AP.
Answer» hence your ap series is -3,1,5,9,13,17............................\xa0 \xa0S10=150 here n=10 ;a=? ;d=?n/2{2a+(n-1)d}150=10/2{2a+9d}30=2a+9deq.1than for next 10th terms we can write S10+S10=150+550S20\xa0=700 than n=20700=20/2{2a+19d}eq.2subtract eq.1 and eq.2 we get\xa040=10dd=4put the value of d in eq.1 we get\xa030=2a+9*4a=-6\xa0\xa0 Ans. Let First Term = a\xa0Common difference = dSum of first 10 Terms S10\xa0= -150\xa0\\(=> S_{10} = {10\\over 2} [2a+ (10-1)d]\\)\\(=> -150 = 5 (2a+9d)\\)=>\xa0\\(2a+9d = -30\\) ........... (1)For Next 10 terms, 11th\xa0ll be first term,\\(=> a_{11} = a + 10d\\)\\(=> S_{10} = {10\\over 2}[2\\times (a+10d) + (10-1)d]\\)\\(=> -550 = 5[2a+20d + 9d]\\)\\(=> 2a+29d = -110 \\) ........ (2)Subtract (1) from (2), we get=> 20d = -80\xa0=> d = -4\xa0Put value of d in(1) we , get\xa0=> a = 3AP : 3 , -1, -5, -9, ........