The sum of the third and the seventh terms of an A.P. is 32 and their product is 220. Find the sum of the first twenty one terms of the A.P. (The terms of the A.P are in ascending order)

Explanation

Let a be the first term and d the common difference of the A.P.
`t_(n)=a+(n-1)d` ……….( Formula)
`t_(3)=a+(3-1)d=a+2d`…………1
`t_(7)=a+(7-1)d=a+6d`…………..2
From the first condition
`a+2d+a+6d=32`…………….[From 1 and 2]
`a+2d+a+6d=32`
`:.2a+8d=32`
`:.a+4d=16`......(Dividing both the sides by 2)
`:.a=16-4d`..............3
From the second condition
`(a+2d)(a+6d)=220` .........[From 1 and 2]
`:.(16-4d+2d)(16-4d+5d)=220`...........[From 3]
`:.(16-2d)(16+2d)=220`
`:.256-4d^(2)=220`
`:.256-220=4d^(2):.4d^(2)=36`
`:.d^(2)=9:.d=+-3`
But the terms are in ascending order.
`:.d=3`
Substituting `d=3` in equation 3
`a+4xx3=16:.a=16-12:.a=4`
Now we want to find the sum of the first 21 terms.
`S_(n)=n/2[2a+(n-1)d]`............(Formula)
`:.S_(21)=21/2[2xx4+(21-1)xx3]` ......(Substituting the values)
`=21/2[8+20xx3]`
`=21/2(8+60)=21/2xx68`
`=21xx34`
`:. S_(21)=714`
Ans. The sum of the first twenty-one terms is 714.