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

(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:

S₇= 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

a₁ = 3 + 4(1)

= 7

n = 2

a₂ = 3 + 4(2)

= 11

d = a₂ - a₁

= 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

a₁ = 9 - 5(1)

= 4

n = 2

a₂ = 9 - 5(2)

= -1

d = a₂ - a₁

= -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

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