Question

16 term of an ap is 5 times the third term if the 10th term is 41 then the first find the sum of first 15 terms

Answer 1

refer the pic for the solution!

Answer 2

Answer:

The sum of 15 term is 495

Step-by-step explanation:

Given that-

The 10th term of A.P = 41

16th term of A.P. = 5×3rd term of A.P.

We know that-

nth term of A.P. is-

$T_{n} = [a + (n -1)d]$

So, 10th term of A.P. is-

$T_{10} = a + (10 - 1)d$

$T_{10} = a + 9d$

41 = a + 9d         --------------------- 1

16th term of A.P. is-

$T_{16} = [a+(16-1)d]$

$T_{16} = a + 15d$          

And, 3rd term of A.P. is-

$T_{3} = [a+(3-1)d]$

$T_{3} = a + 2d$

But, according to question-

$T_{16} = 5T_{3}$

a + 15d = 5[a + 2d]

a + 15d = 5a + 10d

5a - a + 10d - 15d = 0

4a - 5d = 0                    

4a = 5d

$a = \dfrac{5}{4}d$

Put $a = \dfrac{5}{4}d$ in equation 1

$41 = \dfrac{5}{4}d + 9d$

$41 = \dfrac{5d+9d \times4}{4}$

$41 = \dfrac{5d + 36d}{4}$

41×4 = 41d

41d = 164$</p><p> [tex]d = \dfrac{164}{41}$

d = 4

Common difference = 4

Put d = 4 in equation 1

41 = a + 9d

a + 9×4 = 41

a + 36 = 41

a = 41-36

a = 5

First term = 5

We know that-

The sum of n term is-

$S_{n} = \dfrac{n}{2}[2a+(n-1)d]$

So, the sum of first 15 terms is-

$S_{15} = \dfrac{15}{2}[2 \times5 + (15-1) \times4]$

                 = $\dfrac{15}{2}[10+14 \times4]$

                 = $\dfrac{15}{2}[10+56]$

                 = $\dfrac{15}{2} \times66$

                 = 15×33

$S_{n} = 495$

Hence, the sum of 15 term is 495.