# 7 HARD QUESTIONS ON ARITHMETIC PROGRESSION

An arithmetic progression, as I have explained before, is simply a sequence that has the same common difference throughout the sequence.

The nth term of an arithmetic progression can be exemplified as:

$$T_n=a+(n-1)d$$

Where a is the first term

n is the number of terms

d is a common difference.

For the sum of terms of an A.P, the general expression is exemplified as:

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

The above formula is used when the last term is not given. However, when the last term is not given, the formula is as follow:

$$S_n=\frac{n}{2}[a+l]$$

Where l is the last term.

Having said all that, let's now turn our attention to the seven questions.

#### Question 1

Find the first term of an A.P whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

From the first statement,

$16=a+(3-1)d$

$16=a+2d$..... eqn 1

From the second statement,

$T_7=12+T_5$

$a+6d=12+a+4d$

$2d=12$

$d=6$

To obtain the first term, we simply insert 6 in the eqn 1

$16=a+2d$

$16=a+2(6)$

$16=a+12$

$a=4$

#### Question 2

Find the 20th term from the last term of the A.P: 3,8, 13..........253.

Solution:

$d=8-3=5$

The nth term of an A.P from the last term is calculated using:

$$T_n=l-(n-1)d$$

Where l is the last term

Using that same last logic,

$T_{20}=253-(20-1)5$

$T_{20}=253-(19)5$

$T_{20}=253-95=158$

Therefore, the 20th term from the last term is 158

#### Question 3

The sum of the 4th and 8th terms of an A.P is 24 and the sum of the 6th and 10th terms of the A.P is 44. Find the 3rd term of the A.P

Solution:

From the first statement,

$T_4+T_8=24$

$a+3d+a+7d=24$

$2a+10d=24$.....eqn 1

From the second statement,

$T_6+T_10=44$

$a+5d+a+9d=44$

$2a+14d=44$.....eqn 2

Subtracting eqn 2 from Eqn 1

$\begin{matrix} 2a+10d=24\\ -(2a+14d=44)\\ \hline 0-4d=-20 \end{matrix}$

$d=5$

Insert 5 for d in Eqn 1

$2a+10(5)=24$

$2a=24-50$

$2a=-26$

$a=-13$

Since we now know the first term and common difference, we can obtain the first term

$T_3=a+(3-1)d$

$T_3=-13+(2)5$

$T_3=-13+10$

$T_3=-3$

#### Question 4

Jerry started work in 2001 at an annual salary of N5000 and received an increment of N200 each year. In how many years will his salary reach N7000

Solution:

Since he started at N5000, the first term is 5000. N200 increase signifies that the common difference is 200. Hence,

$7000=5000+(n-1)200$

$7000=5000+200n-200$

$7000-5000+200=200n$

$2200=200n$

$n=\frac{2200}{200}=11$

#### Question 5

There are 9 terms in an A.P. The last term of the A.P is 28 and the sum of the terms of the A.P is 144. Find the first term.

Solution:

Just as I told you earlier, the sin of the term of an A.P with a list term is given by

$$S_n=\frac{n}{2}[a+l]$$

By the same reasoning,

$S_9=\frac{9}{2}[a+28]$

Recall that $S_9=144$

$144=\frac{9}{2}[a+28]$

$288=9(a+28)$

$288=9a+252$

$288-252=9a$

$36=9a$

$a=4$

#### Question 6

The first term of an A.P is 155, the last term is 45 and the sum of the terms of the A.P is 400. Find the common difference

Solution:

a=5, l=45 and $S_n=400$

$400=\frac{n}{2}[5+45]$

$400=\frac{n}{2}[5+45]$

$400=\frac{50n}{2}$

$800=50n$

$n=16$

Hence, the sequence has 16 terms.

To obtain the common difference, we simply use the general expression for A.P

$T_n=a+(n-1)d$

$45=5+(16-1)d$

$45=5+15d$

$40=15d$

$d=\frac{40}{15}=\frac{8}{3}$

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#### Question 7

If the ratio of the 11th term to the 18th term of an A.P is 2:3. Find the ratio of the 5th term to the 21st term.

Solution:

Ratio, as I have explained before, is the mother of expression fraction

2:3 is the same as $\frac{2}{3}$, Hence

$frac{T_{11}}{T_{18}}=\frac{2}{3}$

$\frac{a+10d}{a+17d}=\frac{2}{3}$

$3(a+10d)=2(a+17d)$

$3a+30d=2a+34d$

$a=4d$

For the ratio of 5th term to 20th,

$\frac{T_5}{T_{21}}$

$\frac{a+4d}{a+20d}$

But, recalled that $a=4d$, hence, replace a with 4d

$\frac{4d+4d}{4d+20d}=\frac{8d}{24d}=\frac{1}{3}$

Therefore, the ratio of the 5th term to the 21st term is 1:3

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