Question

the sum of the first ten terms of an arithmetic progression consisting of positive integers is equal to the sum of the 20th, 21st and 22nd term. If the first term is less than 20, find how many terms are required to give a sum of 960

Answer 1

Answer:

The correct answer is = 15.

Step-by-step explanation:

Formula:

The sum of the first n terms of an arithmetic progression with first term a and constant difference d is

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

using this formula in this problem

Solution:

The sum of the first ten terms is

$S_{10}=\dfrac{10}{2}[2a+(10-1)d$

$S_{10}=5(2a+9d)$

The sum of the 20th, 21st, and 22nd terms is three times the 21st term:

$3a_{21}=3(a+(21-1)d)$

$3a_{21}=3(a+20d)$

$3a_{21}=3a+60d$

The problem then tells us

$S_{10}=3a_{21}$

$10a+45d=3a+60d$

$7a=15d$

there are only positive integers and the first term a is less than 20 as given. Since 7 and 15 have no common factor, the only explanation of the requirements is a = 15 and d = 7. So the progression is

then, 15, 22, 29, 36, ...

The problem says to find the number of terms n for which the sum is 960:

putting value in the formula

$30n+7n^{2}-7n=1920\\7n^{2}+23n-1920=0$

solving quadratic will give n = 15

thus, the correct answer is 15.