Question

for an ap sum of first 10 and 16 terms are 190 and 400 respectively .find the common difference
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Answer 1

Given,

The sum of first 10 and 16 terms of an AP is 190 and 400 respectively.

To find,

Common difference of the given AP.

Solution,

We can easily solve this mathematical problem by using the following mathematical formula.

Let,

First term of the AP = a

Common difference = d

According to the data mentioned in the question,

number of terms in first AP (n1) = 10

number of terms in second AP (n2) = 16

Now,

Sum of first 10 terms = 10/2 [2a+(10-1)d] = 5×(2a+9d) = 10a+45d

Sum of first 16 terms = 16/2 [2a+(16-1)d] = 8×(2a+15d) = 16a+120d

And,

10a+45d = 190

2a+9d = 38.....(1)

16a+120d = 400

8(2a+15d) = 8×50

2a+15d = 50......(2)

By, subtracting (1) from (2) we get that,

(2a+15d)-(2a+9d) = 50-38

2a+15d-2a-9d = 12

6d = 12

d = 2

Hence, common difference is 2

Answer 2

SOLUTION

GIVEN

For an AP sum of first 10 and 16 terms are 190 and 400 respectively

TO DETERMINE

The common difference

FORMULA TO BE IMPLEMENTED

For an arithmetic progression with

First term = a

Common Difference = d

Sum of first n terms

$\displaystyle \sf{ = \frac{n}{2} \bigg [2a + (n - 1)d \bigg] }$

EVALUATION

Let for the given Arithmetic progression

First term = a

Common Difference = d

Now sum of first 10 terms

$\displaystyle \sf{ = \frac{10}{2} \bigg [2a + (10 - 1)d \bigg] }$

$\displaystyle \sf{ = 5 \bigg [2a + 9d \bigg] }$

By the given condition

$\displaystyle \sf{ 5 \bigg [2a + 9d \bigg] = 190 }$

$\implies \sf{2a + 9d = 38} \: \: .......(1)$

Again sum of 16 terms

$\displaystyle \sf{ = \frac{16}{2} \bigg [2a + (16 - 1)d \bigg] }$

$\displaystyle \sf{ = 8 \bigg [2a + 15d \bigg] }$

By the given condition

$\displaystyle \sf{ 8 \bigg [2a + 15d \bigg] } = 400$

$\implies \displaystyle \sf{ 2a + 15d } = 50 \: \: \: .....(2)$

Now Equation (2) - Equation (1) gives

$\sf{6d = 12}$

$\implies \sf{d = 2}$

FINAL ANSWER

The common difference of the AP = 2

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