Question

The 4th term of an ap is three times the first term and the 7th term exceeds twice the 3rd term by 1.Find the 1st term and the common difference​

Answer 1

Answer:

here's explanation in attachment

Answer 2

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

Let assume that

• First term of an AP = a

• Common difference of an AP = d

Given that,

• The 4th term of an ap is three times the first term.

i.e

$\rm \: a_4 = 3a_1$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.

Tʜᴜs,

$\rm \: a + (4 - 1)d = 3a$

$\rm \: a + 3d = 3a$

$\rm \: 3d = 2a - - - - (1)$

Further given that,

• 7th term exceeds twice the 3rd term by 1.

i.e.

$\rm \: a_7 - 2a_3 = 1$

$\rm \: a + (7 - 1)d - 2[a + (3 - 1)d] = 1$

$\rm \: a + 6d - 2[a + 2d] = 1$

$\rm \: a + 6d - 2a - 4d = 1$

$\rm \: 2d - a = 1$

can be rewritten as on multiply by 2 on both sides

$\rm \: 2 \times (2d - a) = 1 \times 2$

$\rm \: 4d - 2a = 2$

Using equation (1), we get

$\rm \: 4d - 3d = 2$

$\rm\implies \:d = 2$

On substituting d = 2, in equation (1), we get

$\rm \: 2a = 3 \times 2$

$\rm\implies \:\rm \: a = 3$

Hence,

• First term of an AP = 3

• Common difference of an AP = 2

$\rule{190pt}{2pt}$

Additional Information :-

↝ 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.