Question

The 11th term of an progression is 25 and the sum of the first 4 terms is 49. The nth term of the progression is 49

Answer 1

Answer:

For 1: The first term is 10 and the common difference is $\frac{3}{2}$

For 2: The value of n is 27

Step-by-step explanation:

The n-th term of the progression is given as:

$a_n=a_1+(n-1)d$

where,

$a_1$ is the first term, n is the number of terms and d is the common difference

The sum of n-th terms of the progression is given as:

$S_n=\frac{n}{2}[2a_1+(n-1)d]$

where,

$S_n$ is the sum of nth terms

• For (1):

The 11th term of the progression:

$25=a_1+10d$               .......(1)

Sum of first 4 numbers:

$49=\frac{4}{2}[2a_1+3d$              ......(2)

Forming equations:

$98=8a_1+12d$

$25=a_1+10d$                  ( × 8)

The equations become:

$98=8a_1+12d$

$200=8a_1+80d$

Solving above equations, we get:

$102=68d\\\\d=\frac{102}{68}=\frac{3}{2}$

Putting value in equation (1):

$25=a_1+10\frac{3}{2}\\\\a_1=[25-15]=10$

Hence, the first term is 10 and the common difference is $\frac{3}{2}$

• For 2:

The nth term is given as:

$49=10+(n-1)\frac{3}{2}$

Solving the above equation:

$39=(n-1)\frac{3}{2}\\\\n-1=26\\\\n=27$

Hence, the value of n is 27