Sum of 3rd and 7th term of A.P. is 6 and product is 8. Find the sum of First 16 terms.

Sum of 3rd and 7th term of A.P. is 6 and product is 8. Find the sum of

1.	Sum of 3rd and 7th term of A.P. is 6 and product is 8. Find the sum of First 16 terms.
Answer» Here according to question the sum of the 3rd and 7th terms of an A.P. is 6 and their product is 8. We have to\xa0find the sum of the first 20 terms of the A.P.According to question,\xa0a3 + a7= 6 (where a3,a7 represents 3rd\xa0and 7th\xa0terms)And\xa0,a3{tex}\\times{/tex}\xa0a7\xa0=8{tex}{/tex}.Now ,a3= a+2d\xa0and a7= a+6d, are 3rd\xa0\xa0and 7th terms respectively.So,(a+2d)+(a+6d)=6{tex}\\implies{/tex} 2a + 8d = 6{tex}\\implies{/tex}2 (a + 4d)= 6.{tex}\\implies a+4d=\\frac {6 }{2}{/tex}{tex}\\implies a=3-4d{/tex}. And ( a3×a7) = (a + 2d)(a + 6d) = 8.......(i).Substituting the value of a = (3 - 4d) in (i) we get( 3 - 4d + 2d)(3- 4d + 6d) =8{tex}\\Rightarrow{/tex}(3 + 2d)(3 - 2d) = 8{tex}\\Rightarrow{/tex} 9 - 4d2 = 8{tex}\\therefore{/tex}4d2\xa0= 1,d2\xa0=\xa0{tex}\\frac 14{/tex}d =\xa0{tex}\\pm\\frac 12{/tex}Case (i):When\xa0d =\xa0{tex}\\frac 12{/tex}S20\xa0=\xa0{tex}\\frac n2{/tex}[2a+ (n - 1)d]{tex}\\implies{/tex}S20{tex}= \\frac { 20 } { 2 } \\left[ 2 + \\frac { 19 } { 2 } \\right]{/tex}S20\xa0= 115Now, Case(ii):When\xa0d = -{tex}\\frac 12{/tex}S20\xa0=\xa0{tex}\\frac { 20 } { 2 } \\left[ 2 \\times 5 + 19 \\times \\left( - \\frac { 1 } { 2 } \\right) \\right]{/tex}{tex}= 10 \\left[ 10 - \\frac { 19 } { 2 } \\right] {/tex}{tex}=5{/tex}\xa0