Question

6. The fourth, seventh and tenth terms of a G.P. are l,m,n
respectively, then
a) In=m?
b) 12 =m² +n?
c) 12 =mn
d) n² =lm​

Answer 1

Correct Question :-

The fourth, seventh and tenth terms of a G.P. are l,m,n

respectively, then

• a) In=m²

• b) l² =m² + n²

• c) l² =mn

• d) n² =lm

$\large\underline{\sf{Solution-}}$

We know that,

↝ nᵗʰ term of an geometric sequence is,

$\underline{\boxed{ \bf \: a_n \: = \: {ar}^{n - 1}}}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

Tʜᴜs,

↝ 4ᵗʰ term is,

$\rm :\longmapsto\:a_4 \: = \: {ar}^{4 - 1}$

$\rm :\longmapsto\:a_4 \: = \: {ar}^{3}$

➣ It is given that 4ᵗʰ term is l.

$\rm :\longmapsto\: \: \therefore \: \: {ar}^{3} = l - - - (1)$

Aɢᴀɪɴ,

↝ 7ᵗʰ term is,

$\rm :\longmapsto\:a_7 \: = \: {ar}^{7 - 1}$

$\rm :\longmapsto\:a_7 \: = \: {ar}^{6}$

➣ It is given that 7ᵗʰ term is m.

$\rm :\longmapsto\: \: \therefore \: \: {ar}^{6} = m - - - (2)$

Aɢᴀɪɴ,

↝ 10ᵗʰ term is,

$\rm :\longmapsto\:a_{10}\: = \: {ar}^{10 - 1}$

$\rm :\longmapsto\:a_{10}\: = \: {ar}^{9}$

➣ It is given that 10ᵗʰ term is n.

$\rm :\longmapsto\: \: \therefore \: \: {ar}^{9} = n - - - (3)$

Now,

↝ Consider,

$\rm :\longmapsto\:l \times n$

$\: \: \: \: = \: \rm \: \: {ar}^{3} \times {ar}^{9}$

$\: \: \: \: = \: \rm \: \: {a}^{2} {r}^{12}$

$\: \: \: \: = \: \rm \: \: {\bigg( {ar}^{6} \bigg) }^{2}$

$\: \: \: \: = \: \rm \: \: {m}^{2}$

$\bf\implies \:ln = {m}^{2}$

$\bf\implies \: \boxed{ \bf \: Option \: (a) \: is \: correct}$

Additional Information :-

↝ Sum of n terms of an geometric sequence is,

$\underline{ \boxed{ \bf \: S_n = \dfrac{a( {r}^{n} - 1)}{r - 1} \: \: provided \: that \: r \ne \: 1}}$

Wʜᴇʀᴇ,

• Sₙ is the sum of first n terms.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

↝ Sum of infinite terms of an geometric sequence is

$\underline{ \boxed{ \bf \: S_{ \infty} = \dfrac{a}{1 - r} \: \: provided \: that \: |r| < 1 }}$