Question

if for an AP t3=18 and t7=30 then find s17 is equal to​

Answer 1

Answer :

S(17) = 612

Solution :

• Given : t(3) = 18 , t(7) = 30
• To find : S(17) = ?

We know that ,

The nth term of an AP is given by ;

t(n) = a + (n - 1)d , where a is first term and d is common difference .

Thus ,

=> t(3) = a + (3 - 1)d

=> 18 = a + 2d --------(1)

Also ,

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

=> 30 = a + 6d

=> 30 = (a + 2d) + 4d

=> 30 = 18 + 4d { using eq-(1) }

=> 4d = 30 - 18

=> 4d = 12

=> d = 12/4

=> d = 3

Now ,

Putting d = 3 in eq-(1) , we get ;

=> 18 = a + 2d

=> 18 = a + 2•3

=> 18 = a + 6

=> a = 18 - 6

=> a = 12

Now ,

We know that , the sum up to n terms of an AP is given by ; S(n) = (n/2)•[2a + (n - 1)d]

Thus ,

=> S(17) = (17/2)•[2a + (17 -1)•d]

=> S(17) = (17/2)•(2a + 16d)

=> S(17) = (17/2)•(2•12 + 16•3)

=> S(17) = (17/2)•24•(1 + 2)

=> S(17) = 17•12•3

=> S(17) = 612

Hence , S(17) = 612 .