# 7 QUESTIONS ON A.P AND G.P

A Sequence, as I told you previously, is a succession of terms arranged in some definite order such that the terms are related to one another according to a well-defined rule.

So far, we have examined two kinds of sequences.
So, today, we would further our knowledge on A.P and G.P by solving these 7 questions.

Question 1
The fourth them of an A.P is 37 and the 6th term is 12 more than the fourth term. Find the first and seventh terms of the A.P.

Solution:
From the first sentence
$37=a+(4-1)d$
$37=a+3d$.......eqn 1

From the second sentence
$T_6=12+37$
$a+5d=49$.........eqn 2

First, let subtract eqn 2 from 1
$37=a+3d$
$-(49=a+5d)$
$-12=-2d$
$\frac{-12}{-2}=\frac{-2d}{-2}$
$d=6$

Inserting the value of d in eqn 1
$37=a+3(6)$
$37=a+18$
$37-18=a$
$a=19$

Thus, the first term is 19.

To derive the 7th term, we solved using the general term of A.P. in this respect, it will be:
$T_7=19+(7-1)6$
$T_7=19+36$
$T_7=55$

Question 2
The 3rd and 6th terms of a geometric progression are 48 and $\frac{128}{9}$ respectively. Find the 4th term of the G.P.

Solution:
From the question
$48=ar^2$
$\frac{128}{9}=ar^5$

When we divide both eqn, we have
$48÷\frac{128}{9}=r^{2-5}$
$48\times\frac{9}{128}=r^{-3}$
$\frac{432}{128}=r^{-3}$
$\frac{27\times\require{cancel}\bcancel{16}}{8\times\require{cancel}\bcancel{16}}=r^{-3}$
$\frac{27}{8}=r^{-3}$
$\frac{27}{8}=\frac{1}{r^3}$

Now, cross multiply
$27\times r^3=8$

Now divide both side by 27
$\frac{\require{cancel}\bcancel{27}\times r^3}{\require{cancel}\bcancel{27}}=\frac{8}{27}$
$r^3=\frac{8}{27}$

Adding cube root to both sides of the eqn
$\sqrt[3]{r^3}=\sqrt[3]{\frac{8}{27}}$
$r=\frac{2}{3}$

Inserting the value of r in eqn 1
$48=a(\frac{2}{3})^2$
$48=a(\frac{4}{9})$
$48=\frac{4a}{9}$
$48\times9=4a$
$432=4a$

Divide both sides by $4$
$\frac{4a}{4}=\frac{432}{4}$
$a=108$

Now, since the 3rd term is $48$ and common ratio is $\frac{2}{3}$, the fourth term will be $48\times\frac{2}{3}$ which will give $32$.

Recalled the next term of a G.P is derived by multiplying the previous term by the common ratio.

Question 3
The sum of the first and second terms of an A.P is 4 and the 10th term is 19. Find the first and common difference of the A.P.

Solution:
The sum of the first and second terms of the A.P means the sum of the first two terms. Hence,
$4=\frac{2}{2}(2(a)+d)$
$4=2a+d$.....eqn 1

From the second sentence
$19=a+9d$....eqn 2

To solve simultaneously, we multiply Eqn 1 by 1 and eqn 2 by 2.
$1(4=2a+d)$
$2(19=a+9d)$

Substracting the derived eqn
$4=2a+d$
$-(38=2a+18d)$
$-34=-17d$
$\frac{-34}{-17}=\frac{-17d}{-17}$
$d=2$

Inserting the value of d in eqn 1
$4=2a+2$
$4-2=2a$
$2=2a$
$\frac{2a}{2}=\frac{2}{2}$
$a=1$.

Therefore, the first and common difference of the A.P is 1 and 2 respectively.

Question 4
If the second and fifth terms of a geometric progression are $-6$ and $48$ respectively, Find the sum of the first four terms.

Solution:
$-6=ar$
$48=ar^4$

When we divide Eqn 1 by 2, we have:
$-6÷48=r^{1-4}$
$\frac{-6}{48}=r^{-3}$
$\frac{1}{-8}=\frac{1}{r^3}$
$r^3=-8$

Adding square root to both sides
$\sqrt[3]{r^3}=\sqrt[3]{-8}$
$r=-2$

To obtain the first term of the G.P, we insert the value of r in eqn 1
$-6=a(-2)$
$\frac{-6}{-2}=\frac{a\require{cancel}\bcancel{-2}}{\require{cancel}\bcancel{-2}}$
$a=3$

Now, let solve for the sum of the first four terms of the G.P
$S_4=\frac{3(1-(-2)^4)}{1-(-2)}$
$S_4=\frac{\require{cancel}\bcancel{3}(1-16)}{\require{cancel}\bcancel{3}}$
$S_4=1-16$
$S_4=-15$

Question 5
Three numbers x, y, z, in that order, are in geometric progression. Given further that x, 2y, z in that order are in arithmetic progression. Determine the possible values of the common ratio of the G.P.

Solution:
Since they are in geometric progression,
$x=a$
$y=ar$
$z=ar^2$

Since it is an A.P, we can simply derive the
common difference by subtracting consecutive terms of an A.P.
$2y-x=z-2y$

By substitution
$2(ar)-a=ar^2-2(ar)$
$2ar-a=ar^2-2ar$
$2ar+2ar=ar^2+a$
$4ar=ar^2+a$

By dividing through by $a$, we have:
$4r=r^2+1$
$0=r^2-4r+1$
$r^2-4r+1=0$

Using the quadratic formula
$a=1$, $b=-4$, $C=1$
$x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)(1)}$
$r=\frac{+4\pm\sqrt{16-4}}{2}$
$r=\frac{+4\pm\sqrt{12}}{2}$
$r=\frac{+4\pm\sqrt{4\times3}}{2}$
$r=\frac{+4\pm2\sqrt{3}}{2}$
$r=\frac{+4+2\sqrt{3}}{2}$ or $\frac{+4-2\sqrt{3}}{2}$

$2$ is common to both sides, hence
$r=2+\sqrt{3}$  or  $2-\sqrt{3}$

Thus, the possible values of $r$ are $2+\sqrt{3}$ and $2-\sqrt{3}$

Example 6
The common ratio of a G.P is $2$. If the 5th term is greater than the 1st term by 45. Find the fifth term of the G.P

Solution:
$T_5=45+a$

If $T_5=a(2)^{5-1}$, then
$a(2)^{5-1)}=45+a$
$a(2)^{4}=45+a$
$a(16)=45+a$
$16a-a=45$
$15a=45$
$\frac{15a}{15}=\frac{45}{15}$
$a=3$

Substituting 3 for $a$ in the first eqn
$T_5=45+3$
$T_5=48$

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Example 7
The third term of the G.P is $360$ and the sixth term is $1215$. Find the
1. Common ration
2. First term
3. Sum of the first four terms.

Solution:
$360=ar^2$......eqn 1
$1215=ar^5$......eqn 2

Solving simultaneously by dividing both eqns
$\frac{360}{1215}=r^{2-5}$
$\frac{8\times\require{cancel}\bcancel{45}}{27\times\require{cancel}\bcancel{45}}=r^{-3}$
$\frac{8}{27}=\frac{1}{r^3}$

Cross multiply
$8\times r^3=27$

Divide both sides by $8$
$\frac{\require{cancel}\bcancel{8}\times r^3}{\require{cancel}\bcancel{8}}=\frac{27}{8}$
$r^3=\frac{27}{8}$

Add cube root to both sides
$\sqrt[3]{r^3}=\sqrt[3]{\frac{27}{8}}$
$r=\frac{3}{2}$

Now, inserting the value of $r$ in eqn 1
$360=a(\frac{3}{2})^2$
$360=a(\frac{9}{4})$
$360=\frac{9a}{4}$

After cross multiplying, we would have:
$1440=9a$
$\frac{1440}{9}=\frac{9a}{9}$
$a=160$

Now, to the sum of the first four terms of the G.P
$S_4=\frac{160(\frac{3}{2})^4-1)}{\frac{3}{2}-1}$
$S_4=\frac{160(\frac{81}{16}-1)}{\frac{3-2}{2}}$
$S_4=\frac{160(\frac{81-16}{16})}{\frac{1}{2}}$
$S_4=\frac{160(\frac{65}{16})}{\frac{1}{2}}$
$S_4=\frac{10400}{16}÷\frac{1}{2}$
$S_4=650÷\frac{1}{2}$

To change division to multiplication, we overt the fraction
$S_4=650\times2$
$S_4=1300$
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