Answer :

First term , a = 18

S(20) = 740

Note :

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

★ If a1 , a2 , a3 , . . . , an are in AP , then

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ The common difference of an AP is given by ; d = a(n) - a(n-1) .

★ The NTH term of an AP is given by ;

a(n) = a + (n - 1)d .

★ If a , b , c are in AP , then 2b = a + c .

★ The sum of nth terms of an AP is given by ; S(n) = (n/2)×[ 2a + (n - 1)d ] .

or S(n) = (n/2)×(a + l) , l is the last term .

Solution :

• Given : a(7) = 30 , a(17) = 50
• To find : First term , a = ? , S(20) = ?

We know that , the nth term of an AP is given by ;

a(n) = a + (n - 1)d .

Thus ,

=> S(7) = a + (7 - 1)d

=> 30 = a + 6d

=> a = 30 - 6d ------(1)

Also ,

=> S(17) = a + (17 - 1)d

=> 50 = a + 16d

=> a = 50 - 16d ------(2)

From eq-(1) and (2) , we have ;

=> 30 - 6d = 50 - 16d

=> 16d - 6d = 50 - 30

=> 10D = 20

=> d = 20/10

=> d = 2

Now ,

Putting d = 2 in eq-(1) , we get ;

=> a = 30 - 6d

=> a = 30 - 6•2

=> a = 30 - 12

=> a = 18

Hence ,

The first term of the AP is 18 .

Also ,

We know that , the sum of first n terms of an AP is given by ;

S(n) = (n/2)•[ 2a + (n - 1)d ]

Thus ,

The sum of first 20 terms of the AP will be given as ;

=> S(20) = (20/2)•[ 2a + (20 - 1)d ]

=> S(20) = 10•[ 2a + 19d ]

=> S(20) = 10•[ 2•18 + 19•2 ]

=> S(20) = 10•[ 36 + 38 ]

=> S(20) = 10•74

=> S(20) = 740

Hence S(20) = 740 .