The sum of the third and the seventh terms
of an AP is 6 and their product is 8. Find
the sum of first sixteen terms of the AP.
Answers
Answer :
Sum of first sixteen terms of an AP is 76.
Explanation :
Given that,
a3 + a7 = 6
=> a + 2d + a + 6d = 6
=> 2a + 8d = 6
=> a + 4d = 3
=> a = 3 - 4d ______________________(1)
It is also given that,
a3 × a7 = 8
=> (a+2d)(a+6d) = 8______________(2)
Putting the value of (1) in (2),
=> (3-2d)(3+2d) = 8
=> 9 - 4d² = 8
=> -4d² = -1
=> d = 1/2
Put value of d in (1),
=> a = 3 - 4(1/2)
=> a = 3 - 2
=> a = 1
We know that,
Sn = n/2[2a+(n-1)d]
We have to calculate S16,
=> S16 = 16/2[2(1)+(15)(1/2)]
=> S16 = 8(2+15/2)
=> S16 = 8(19/2)
=> S16 = 4 × 19
=> S16 = 76
∴ Sum of first sixteen terms of an AP is 76.
Given :-
→ a₃ + a₇ = 6.
→ a₃ × a₇ = 8 .
To find :-
→ S₁₅ .
Solution :-
We have,
→ a₃ + a₇ = 6.
⇒ a + 2d + a + 6d = 6 .
⇒ 2a + 8d = 6 .
⇒ 2( a + 4d ) = 6 .
⇒ a + 4d = 6/2 .
⇒ a + 4d = 3
∵a = 3 - 4d ............(1) .
And,
→ a₃ × a₇ = 8 .
⇒ ( a + 2d ) × ( a + 6d ) = 8 .
⇒ ( 3 - 4d + 2d ) ( 3 - 4d + 6d ) = 8 .
⇒ ( 3 - 2d )( 3 + 2d ) = 8 .
⇒ 3² - (2d)² = 8 .
⇒ 9 - 4d² = 8.
⇒ 4d² = 9 - 8 .
⇒ 4d² = 1 .
⇒ d² = 1/4 .
⇒ d = √(1/4) .
∴ d = 1/2 .
Putting the value of d in equation (1), we get ,
⇒ a = 3 - 4 × 1/2 .
⇒ a = 3 - 2 .
∴ a = 1 .
Thus, sum of 16th term is given by ,
S16 =76
Hence, sum of first 16 terms is 76.