Yhe sum of the third term and seventh term of ap is 6 and product is 8 . Find sum of sixteen terms
Answers
the sum of 3rd term and 7th term = a+2d+a+6d = 6, or
2a+8d = 6, or
a + 4d = 3 …(1)
Their product is (a+2d)*(a+6d) = 8 …(2)
From (1), a = 3–4d. Put that in (2) to get
(a+2d)*(a+6d) = 8, or
(3–4d+2d)*(3–4d+6d) = 8, or
(3–2d)*(3+2d) = 8, or
9–4d^2 = 8, or
4d^2 = 1, or
d^2 = 1/4, or
d = 1/2.
From (1), a = 3–4d = 3–2 = 1.
The sum of the first 16 terms
S16 = (n/2)[2a+(n-1)d]
= (16/2)[2*1 + (16–1)*1/2]
= 8[2+15/2]
= 8 *19/2
= 76. Answer.
Answer:
Step-by-step explanation:
Let the first term be a and common difference be d
nth term = a+(n-1)d
Given Third Term + Seventh term = (a+2d)+(a+6d) = 6 ==> a+4d = 3
hence, a= 3-4d
Third Term * Seventh term = (a+2d)*(a+6d) = 8
(3-4d+2d)*(3-4d+6d) = 8==> (3-2d)*(3+2d) = 8
i.e. 9-4d^2 = 8==> d^2 = (9-8)/4 = 0.25==> d = 0.5 or -0.5
Now to check which is correct d...
Substitute and find
Case (a): d= 0.5
a+4d = 3==> a=3-4d = 3-4(0.5)=1
3rd term = a+2d= 1+2*0.5 = 2
7th term = a+6d= 1+6*0.5 = 4
Sum = 6 and Product = 8
Case (b): d= -0.5
a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 5
3rd term = a+2d= 5+2*(-0.5) = 4
7th term = a+6d= 5+6*(-0.5) = 2
Sum = 6 and Product = 8
Since both are matching, we will go with bothvalues
Sum of first 16 terms = n*(2a+(n-1)d)/2 = 16*(2a+15d)/2
= 8*(2a+15d)
Case (a): d= 0.5
Sum = 8*(2*1+15*0.5)=76
Case (b): d= 0.5
Sum = 8*(2*5+15*(-0.5))=20