The sum of 3rd term 7th term is 6 and their product is 8 find tge sum of 1st 6th term
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tn= a +(n-1)d
By given condition,
t3 + t7 =6. ..............(i)
t3 × t7 =8. ................(ii)
t3 = a+2d
t7 = a+6d
Putting in eq (i) & (ii)
a+2d +a+6d =6
2a +8d =6
a+4d =3
a=3-4d
a+2d ×(a+6d) =8
a(a+6d) +2d(a+6d) =8
a^2 +6ad +2ad + 12d^2 = 8
a^2 +8ad +12d^2 = 8
putting a=3-4d
(3-4d)^2 +8ad +12 d^2 = 8
3^2 -2×3×-4d +4d^2 +8ad + 12d^2 = 8
9-24d+16d^2 +8ad +144d^2 =8
By given condition,
t3 + t7 =6. ..............(i)
t3 × t7 =8. ................(ii)
t3 = a+2d
t7 = a+6d
Putting in eq (i) & (ii)
a+2d +a+6d =6
2a +8d =6
a+4d =3
a=3-4d
a+2d ×(a+6d) =8
a(a+6d) +2d(a+6d) =8
a^2 +6ad +2ad + 12d^2 = 8
a^2 +8ad +12d^2 = 8
putting a=3-4d
(3-4d)^2 +8ad +12 d^2 = 8
3^2 -2×3×-4d +4d^2 +8ad + 12d^2 = 8
9-24d+16d^2 +8ad +144d^2 =8
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