HOW TO FIND THE SUM OF SERIES IN AP WITH GIVEN INFORMATION

Question 1 :

The first and last term of an AP are 17 and 350 respectively.If the common difference is 9, how many terms are there and what is their sum?

Solution :

a = 17       l = 350   d = 9  

S_n  =  (n/2) [2a + (n-1)d]

S_n  =  (n/2) [2(17)+(n-1)9]

  =  (n/2) [34 + 9 n - 9]

  =  (n/2) [25 + 9 n]  ----(1)

S_n  =  (n/2)[a + l]

  =  (n/2)[17 + 350] ------(2)

(1)  =  (2)

(n/2) [25 + 9 n]  =  (n/2)[17 + 350]

25 + 9 n = 17 + 350

9n  =  367 - 25

  9n  =  342

 n = 342/9

  n  = 38 

38 terms are needed.

S₃₈ = (38/2)[17 + 350]

  =  19 [367]

  =  6973

Question 2 :

The sum of first 22 terms of an AP in which d = 7and 22nd term is 149.

Solution :

n = 22  d = 7 

a₂₂  =  a + 21 d

  =  a + 21 (7)

 149 = a + 147

 149 -147 = a

  a = 2

 S_n  =  (n/2) [2a + (n-1)d]

  =  (22/2) [2(2) + (22-1)7]

  =  11 [4 + 21(7)]

  =  11 [4 + 147]

  =  11 [151]

  =  1661

Question 3 :

The sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution :

a₂ = a + 1d

14 = a + d

 a + d = 14 ---(1)

 a₃ = a + 2d

 18 = a + 2d 

 a + 2 d = 18 ------(2)

 (1) - (2)

       a + d = 14

       a + 2 d = 18

     (-)   (-)   (-)

     --------------

            - d = -4

              d = 4

By applying the value of d = 4 in the (1), we get

a + 4  =  14

a = 14 - 4

 a = 10

S_n = (n/2) [2a + (n - 1) d]

 n = 51

S_n  =  (51/2)[2(10) + (51-1) 4]

  =  (51/2)[20 + 50(4)]

  =  (51/2) [20 + 200]

  =  (51/2) [220]

  =  51 (110)

S₅₁  =  5610

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