Question

The sum of first 6 term of

a.p is 42. the ratio its 10th term to 38th term is 1:3. calculate the first and 13th term of the

a.p

Answer 1

Answer:

The first term of the a.p is $a_{1} = 4.675$ and the 13th term of the ap is 15.835 .

Step-by-step explanation:

As the airthmetic series is given by

$a_{n} = a_{1}+ (n - 1)d$

$S_{n} = \frac{n}{2}(a_{1}+a_{n})$

Where $a_{n}$ is the nth term , $a_{1}$ is the first term and d is the common difference .

As given

The sum of first 6 term of  a.p is 42.

$a_{6} = a_{1}+ (6 - 1)d$

$a_{6}= a_{1}+5d$

$S_{6} = \frac{6}{2}[a_{1}+a_{1}+5d]$

$42 = 3[2a_{1}+5d]$

$\frac{42}{3} = 2a_{1}+5d$

$14= 2a_{1}+5d$

The ratio its 10th term to 38th term is 1:3.

$\frac{a_{1}+(10-1)d}{a_{1}+(38-1)d} = \frac{1}{3}$

$3\times {a_{1}+(10-1)d}={a_{1}+(38-1)d}$

${3a_{1}+3\times 9d}={a_{1}+(38-1)d}$

${3a_{1}+27d}={a_{1}+37d}$

${3a_{1}-a_{1}+27d-37d=0$

$2a_{1}-10d=0$

$a_{1}-5d=0$

Two Equation becomes

$14= 2a_{1}+5d$

$a_{1}-5d=0$

Mulitply $a_{1}-5d=0$ by 2 and subtracted from $14= 2a_{1}+5d$ .

$2a_{1}-2a_{1}+5d+10d = 14$

$15d = 14$

$d = \frac{14}{15}$

d = 0.93 (Approx)

Put in the equation

$14= 2a_{1}+5\times 0.93$

$14= 2a_{1}+4.65$

$14-4.65= 2a_{1}$

$9.35= 2a_{1}$

$a_{1} = \frac{9.35}{2}$

$a_{1} = 4.675$

Thus the first term is 4.675 .

For 13th term

$a_{13} =4.675+ (13 - 1)\times 0.93$

$a_{13} =4.675+12\times 0.93$

$a_{13} =4.675+11.16$

$a_{13} =15.835$

Therefore the first term of the a.p is $a_{1} = 4.675$ and the 13th term of the ap is 15.835 .

Answer 2

Answer:A=14/3

A13=238/5