ii) The sum of the 3rd and 7th terms of an A.P. is 32 and their product is 220. Find
the sum of first twenty-one terms of the A.P. (Take the value of d positive).
Answers
Answer:
a+2d+a+6d= 32
2a+8d= 32
a+4d= 16
To find= 21/2 (a+20d)
= 2(a+10d)=?
(a+2d) (a+6d) = 220
(16-4d+zd) (16-4d+6d) = 220
(16-2d) (16+2d) = 220
256-4d²= 220
36/4 = d²
So, d= 3, a = 4
21(4+30)= 21×34
Ans = 714
Solution :-
The sum of 3rd and 7th term of an AP is 32 .
Let the first term and common difference be a and d
Therefore,
According to the question,
a + 2d + a + 6d = 32
2a + 8d = 32
2(a + 4d) = 32
a + 4d = 32/2
a + 4d = 16
a = 16 - 4d. eqn( 1 )
Now,
Product of 3rd and 7th term is 220
Therefore,
( a + 2d) ( a + 6d) = 220
Subsitute the value of a that is eqn ( 1 )
( 16 - 4d + 2d) ( 16 - 4d + 6d) = 220
( 16 - 2d) ( 16 + 2d) = 220
As we know that,
[ ( a + b) ( a - b) = a^2 - b^2]
( 16)^2 - ( 2d)^2 = 220
256 - 4d^2 = 220
-4d^2 = 220 - 256
-4d^2 = -36
d^2 = -36/-4
d^2 = 9
d = √9
d = 3
Substitute the value of d in eqn(1)
a = 16 - 4 * 3
a = 16 - 12
a = 4
Now, we have to find the sum of first 21 terms of an AP.
Therefore,
By using formula to find the sum of terms of an AP.
Sn = n/2 ( 2a + ( n - 1 )d)
S21 = 21/2 ( 2 * 4 + ( 21 - 1 ) 3 )
S21 = 21/2 ( 8 + 60 )
S21 = 21/ 2 * 68
S21 = 21 * 34
S21 = 714
Hence, The sum of first 21 terms of an AP is 714 
Formula kept in mind :-
• an = a + ( n - 1 )d
• Sn = n/2 ( 2a + (n - 1 )d)