If 14th term of an AP is twice its 8th term is -8, then find the sum of its first 20 terms.

If 14th term of an AP is twice its 8th term is -8, then find the sum o

1.	If 14th term of an AP is twice its 8th term is -8, then find the sum of its first 20 terms.
Answer» Let first term be a and common difference be d.Here, a14 = 2a8a + 13d = 2(a + 7d)a + 13d = 2a + 14da = - d...(i)a\u200b\u200b\u200b\u200b\u200b\u200b6\xa0= - 8a + 5d = -8 ... (ii)Putting the value of a from (i) in (ii), we get-d + 5d = -84d = -8d = -2Put d = -2 in (i)a = -(-2)a = 2So ,a = 2, d = - 2{tex}S _ { 20 } = \\frac { 20 } { 2 } [ 2 \\times 2 + ( 20 - 1 ) ( - 2 ) ]{/tex}{tex}= 10 [ 4 + 19 \\times ( - 2 ) ]{/tex}= 10(4 - 38)= 10\xa0{tex}\\times{/tex}(-34)= - 340. Which is the required sum of first 20 terms.