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Find the Sum of First 51 Terms of an Ap Whose Second and Third Terms Are 14 and 18 Respectively. - CBSE Class 10 - Mathematics

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Question

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

Solution 1

Given that,

a2 = 14

a3 = 18

d = a3 − a2 = 18 − 14 = 4

a2 = a + d

14 = a + 4

a = 10

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

`S_51 = 51/2[2xx10+(51-1)4]`

`=51/2[20+(50)(4)]`

`=(51(220))/2=51(110)`

= 5610

Solution 2

In the given problem, let us take the first term as a and the common difference as d

Here, we are given that,

`a_2 = 14` .....(1)

`a_3 = 18` ....(2)

Also we know

`a_n = a + (n -1)d`

For the 2nd term (n = 2)

`a_2 = a + (2 - 1)d`

14 = a + d    (Using 1)

a = 14 - d   .....(3)

Similarly for the 3rd term (n = 3)

`a_3 = a + (3 - 1)d`

18 = a + 2d     (Using 2)

a = 18 - 2d .......(4)

Subtracting (3) from (4), we get,

a - a = (18 - 2d) - (14 - d)

0 = 18 - 2d - 14 + d

0 = 4 - d

d = 4

Now, to find a, we substitute the value of d in (4),

a = 14 - 4

a = 10

So for the given A.P d = 4 and a = 10

So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

So, using the formula for n = 51, we get,

`S_51 = 51/2 [2(10) + (51 - 1)(4)]`

`= 51/2 [20 + (50)(4)]`

`= 51/2 [20 + 200]`

`= 51/2 [220]`

= 5610

Therefore the usm of first 51 terms for the given A.P is `S_51 = 5610`

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APPEARS IN

 NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 5: Arithmetic Progressions
Ex. 5.30 | Q: 8 | Page no. 113
Solution Find the Sum of First 51 Terms of an Ap Whose Second and Third Terms Are 14 and 18 Respectively. Concept: Sum of First n Terms of an AP.
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