Question

the sum of the first 7 terms of an arithmetic progression is 140 and the sum of the next 7 terms of the same progression is 385 then find the arithmetic progression​

Answer 1

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

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.

Let's solve the problem now!!

↝ Let assume that

• First term of an AP is a

• Common difference of an AP is d.

According to statement,

↝ Sum of first 7 terms of anAP = 140.

$\rm :\longmapsto\:S_7 = 140$

$\rm :\longmapsto\:\dfrac{7}{2}\bigg(2a + (7 - 1)d\bigg) = 140$

$\rm :\longmapsto\:2a + 6d = 40$

$\rm :\longmapsto\:a + 3d = 20 - - - (1)$

According to statement again,

↝ Sum of next 7 terms be 385.

Since, Sum of first 7 terms is 140.

So, it implies,

↝ Sum of first fourteen terms is 525.

$\rm :\longmapsto\:S_{14} = 525$

$\rm :\longmapsto\:\dfrac{14}{2}\bigg(2a + (14 - 1)d\bigg) = 525$

$\rm :\longmapsto\:2a + 13d = 75$

$\rm :\longmapsto\:2(20 - 3d) + 13d = 75 \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:40 - 6d + 13d = 75$

$\rm :\longmapsto\:7d = 75 - 40$

$\rm :\longmapsto\:7d = 35$

$\bf\implies \:d = 5$

On substituting the value of d, in equation (1), we get

$\rm :\longmapsto\:a + 3 \times 5 = 20$

$\rm :\longmapsto\:a + 15 = 20$

$\bf\implies \:a = 5$

Hence,

$\: \: \: \: \: \underbrace{ \boxed{ \bf{ \: AP \: series \: is \: 5, \: 10, \: 15, \: 20, - - - - }}}$

Additional Information :-

↝ 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.