In an increasing G.P. The sum of the first and the last term is 66, the product of the second and the last but one term is 128, and the sum of all the terms is 126. How are terms in progression?
Last updated date: 23rd Mar 2023
•
Total views: 306.3k
•
Views today: 2.84k
Answer
306.3k+ views
Hint: Use the nth term formula and proceed as instructed in question.
Complete step-by-step answer:
It has been given
$
a + a{r^{n - 1}} = 66 \\
{\text{And }} \\
ar.{a^{n - 2}} = {a^2}{r^{n - 1}} = 128 \\
\therefore {a^{n - 1}} = \dfrac{{128}}{a} \\
{\text{Putting in 1, we get }}a + \dfrac{{128}}{a} = 66 \\
\therefore {a^2} - 66a + 128 = 0 \\
{\text{splitting the middle terms we get}} \\
(a - 2)(a - 64) = 0 \\
\therefore a = 2,64 \\
{r^{n - 1}} = 32,\dfrac{1}{{32}} \\
{\text{We reject the second value as r > 1 ,}}\therefore {{\text{r}}^{n - 1}} = 32 \\
{\text{Sum = }}\dfrac{{a({r^n} - 1)}}{{r - 1}} = 126{\text{ }} \Rightarrow \dfrac{{2(32r - 1)}}{{r - 1}} = 126 \\
\because {r^{n - 1}} = 32 \\
\therefore 32r - 1 = 63r - 63 \\
\therefore r = 2{\text{ and }}{r^{n - 1}} = 32{\text{ gives}} \\
{{\text{2}}^{n - 1}} = {2^5} \Rightarrow n - 1 = 5 \Rightarrow n = 6 \\
$
Note: Must remember all the general terms, sum formula, quadratic roots in order to solve such similar problems. Similar questions can be asked for Harmonic Progression and AGP etc.
Complete step-by-step answer:
It has been given
$
a + a{r^{n - 1}} = 66 \\
{\text{And }} \\
ar.{a^{n - 2}} = {a^2}{r^{n - 1}} = 128 \\
\therefore {a^{n - 1}} = \dfrac{{128}}{a} \\
{\text{Putting in 1, we get }}a + \dfrac{{128}}{a} = 66 \\
\therefore {a^2} - 66a + 128 = 0 \\
{\text{splitting the middle terms we get}} \\
(a - 2)(a - 64) = 0 \\
\therefore a = 2,64 \\
{r^{n - 1}} = 32,\dfrac{1}{{32}} \\
{\text{We reject the second value as r > 1 ,}}\therefore {{\text{r}}^{n - 1}} = 32 \\
{\text{Sum = }}\dfrac{{a({r^n} - 1)}}{{r - 1}} = 126{\text{ }} \Rightarrow \dfrac{{2(32r - 1)}}{{r - 1}} = 126 \\
\because {r^{n - 1}} = 32 \\
\therefore 32r - 1 = 63r - 63 \\
\therefore r = 2{\text{ and }}{r^{n - 1}} = 32{\text{ gives}} \\
{{\text{2}}^{n - 1}} = {2^5} \Rightarrow n - 1 = 5 \Rightarrow n = 6 \\
$
Note: Must remember all the general terms, sum formula, quadratic roots in order to solve such similar problems. Similar questions can be asked for Harmonic Progression and AGP etc.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
