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Find the Sum 3 + 11 + 19 + ... + 803

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Question

Find the sum 3 + 11 + 19 + ... + 803

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Solution

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here 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

3 + 11 + 19 + ... + 803

Common difference of the A.P. (d) = `a_2 - a_1`

= 19 - 11

= 8

So here,

First term (a) = 3

Last term (l) = 803

Common difference (d) = 8

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

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

So, for the last term,

Further simplifying,

803 = 3 + (n -1)8

803 = 3 + 8n - 8

803 + 5 = 8n

808 = 8n

`n = 808/8`

n = 101

Now, using the formula for the sum of n terms, we get

`S_n = 101/2[2(3) + (101 - 1)8]`

`= 101/2 [6 + (100)8]`

`= 101/2 (806)`

= 101(403)

= 40703

Therefore, the sum of the A.P is `S_n = 40703`

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Chapter 5: Arithmetic Progressions - Exercise 5.6 [Page 51]

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R.D. Sharma Mathematics [English] Class 10
Chapter 5 Arithmetic Progressions
Exercise 5.6 | Q 13.2 | Page 51

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