Question

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 :​

Answer 1

Answer:

Step-by-step explanation:

(25/2)(2a+24d) = (15/2)(2(a+25d)+14d)

50a+600d = 15[2a+50d+14d]

20a+600d = 960d

60 = 360d

d = 1/6

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

First term of an AP series, a = 3

Let assume that

Common difference of an AP series is 'd'.

Further given that,

➢ Sum of its first 25 terms is equals to sum of its next 15 terms.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

So, According to statement

$\rm :\longmapsto\:S_{25} = S_{40} - S_{25}$

$\rm :\longmapsto\:2S_{25} = S_{40}$

$\rm :\longmapsto\:2 \times \dfrac{25}{2} \bigg(2a + (25 - 1)d \bigg) = \dfrac{40}{2} \bigg(2a + (40 - 1)d \bigg)$

$\rm :\longmapsto\:25 \bigg(2(3) + 24d \bigg) = 20 \bigg(2(3) + 39d \bigg)$

$\rm :\longmapsto\:5\bigg(6 + 24d \bigg) = 4 \bigg(6 + 39d \bigg)$

$\rm :\longmapsto\:30 + 120d = 24 + 156d$

$\rm :\longmapsto\:30 - 24 =156d - 120d$

$\rm :\longmapsto\:6 =36d$

$\bf\implies \:d \: = \: \dfrac{1}{6}$

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↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.