The question of arithmetic progressions of exercise 5.4 of class 10th math of optional chapter, sir please help me for solving this question 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.

AnilSinghBoraGuru

# 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. Q.2

Share

From the given statements, we can write,

a

_{3 + }a_{7}= 6 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)And

a

_{3 }Ã—a_{7 }= 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)By the nth term formula,

aÂ =Â_{n}a+(nâˆ’1)dThird term,a_{3 }= a+(3 -1)da

_{3 }= a + 2dâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iii)And Seventh term, a7= a+(7-1)d

a

_{7 }= a + 6d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iv)From equation (iii) and (iv), putting in equation(i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

or

a = 3â€“4d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)Ã—(a+6d) = 8

Putting the value of a from equation (v), we get,

(3â€“4d +2d)Ã—(3â€“4d+6d) = 8

(3 â€“2d)Ã—(3+2d) = 8

3

^{2 }â€“ 2d^{2}= 89 â€“ 4d

^{2}= 84d

^{2}= 1d = 1/2 or -1/2

Now, by putting both the values of d, we get,

a = 3 â€“ 4d = 3 â€“ 4(1/2) = 3 â€“ 2 = 1, when d = 1/2

a = 3 â€“ 4d = 3 â€“ 4(-1/2) = 3+2 = 5, when d = -1/2

We know, the sum of nth term of AP is;

SÂ =Â_{n}n/2Â [2aÂ +(nÂ â€“ 1)d]So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

SÂ =Â_{16}16/2Â [2Â +(16-1)1/2] = 8(2+15/2) = 76And when a = 5 and d= -1/2

Then, the sum of first 16 terms are;

SÂ =Â_{16}16/2Â [2.5+(16-1)(-1/2)] = 8(5/2)=20