10th CBSE math solution for exercise 5.3 part 7

This page 10th CBSE math solution for exercise 5.3 part 7 is going to provide you solution for every problems that you find in the exercise no 5.3 part 7

10th CBSE math solution for exercise 5.3 part 7

(9) If the sum of 7 terms of an AP is 49 and that of 17 terms is 289,find the sum of first n terms.

Solution:

S= 49

S₁₇= 289

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

S₇= (7/2) [2 a + (7 - 1) d]

49 = (7/2) [ 2a + 6d]

(49 x 2)/7 = 2 a + 6 d

  14 = 2 a + 6 d

  a + 3 d = 7 -----(1)

 S₁₇= (17/2) [2 a + (17 - 1) d]

289 = (17/2) [ 2a + 16d]

(289 x 2)/17 = 2 a + 16 d

  34 = 2 a + 16 d

  a + 8 d = 17 -----(1)

(1) - (2)

           a + 3 d = 7

        a + 8 d = 17

       (-)  (-)    (-)

       -------------

            - 5 d = -10

                 d = 2

 Substitute d = 2 in the first equation

    a + 3 (2) = 7

   a + 6 = 7

       a = 7 - 6

        a = 1

To find the sum of first n terms, we have to apply the values of a and in the Sn formula

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

 S n = (n/2) [2(1) + (n - 1) (2)]

       = (n/2) [2 + 2 n - 2]

       = (n/2) [2 n]

       = n²


(10) Show that a₁,a₂,............ an form an AP where an is defined as below

(i) a n = 3 + 4 n

(ii) a n = 9 - 5 n

Also find the sum of 15 terms in each case.

Solution:

(i) a n = 3 + 4 n

 n = 1

 a= 3 + 4(1)

     = 7

n = 2

a₂  = 3 + 4(2)

     = 11

 d = a₂ -  a

     11 - 7

    = 4 

So the series will be in the form 7 + 11 + ...........

Now we need to find sum of 15 terms

 S₁₅ = (n/2) [ 2 a + (n - 1) d]

      = (15/2) [2(7) + (15-1) 4]

      = (15/2) [14 + 14(4)]

      = (15/2) [14 + 56]

      = (15/2) [70]

      = (15 x 35)

      = 525


In the page 10th CBSE math solution for exercise 5.3 part 7 next we are going to see the solution of next problem.

(ii) a n = 9 - 5 n

Solution:

n = 1

 a= 9 - 5(1)

     = 4

n = 2

a₂  = 9 - 5(2)

     = -1

 d = a₂ -  a

     -1 - 4

    = -5

So the series will be in the form 4 + (-1) + ...........

Now we need to find sum of 15 terms

 S₁₅ = (n/2) [ 2 a + (n - 1) d]

      = (15/2) [2(4) + (15-1) (-5)]

      = (15/2) [8 + 14(-5)]

      = (15/2) [8 - 70]

      = (15/2) [-62]

      = 15 x (-31)

      = -465

       back to exercise




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