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.
