JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 16)
If the first term of an A.P. is 3 and the sum of
its first 25 terms is equal to the sum of its next
15 terms, then the common difference of this
A.P. is :
$${1 \over 4}$$
$${1 \over 5}$$
$${1 \over 7}$$
$${1 \over 6}$$
Explanation
First 25 terms = a, a + d, .......,a + 24d
Next 15 terms = a + 25d, a + 26d, ......, a + 39d
$$ \therefore $$ $${{25} \over 2}\left[ {2a + 24d} \right] = {{15} \over 2}\left[ {2\left( {a + 25d} \right) + 14d} \right]$$
$$ \Rightarrow $$ 50a + 600d = 15 [2a + 50d + 14d]
$$ \Rightarrow $$ 20a + 600d = 960d
$$ \Rightarrow $$ 60 = 360d
$$ \Rightarrow $$ d = $${1 \over 6}$$
Next 15 terms = a + 25d, a + 26d, ......, a + 39d
$$ \therefore $$ $${{25} \over 2}\left[ {2a + 24d} \right] = {{15} \over 2}\left[ {2\left( {a + 25d} \right) + 14d} \right]$$
$$ \Rightarrow $$ 50a + 600d = 15 [2a + 50d + 14d]
$$ \Rightarrow $$ 20a + 600d = 960d
$$ \Rightarrow $$ 60 = 360d
$$ \Rightarrow $$ d = $${1 \over 6}$$
Comments (0)
