Question

the fourth, seventh and tenth term of G.P are p, q, r respt. then prove that q^2 = pr​

Answer 1

GIVEN :–

• Fourth term of G.P. = p

• Seventh term of G.P. = q

• Tenth term of G.P. = r

TO PROVE :–

• q² = pr

SOLUTION :–

• We know that nth term of G.P. is –

$\\ \longrightarrow \large{ \boxed{ \bold{ T_{n} = a {r}^{n - 1} }}}$

• Here –

$\\ \: \: \: \: \to \: \: \: { \bold{a = first \: \: term}}$

$\\ \: \: \: \: \to \: \: \: { \bold{r = common \: \: ratio}}$

$\\ \: \: \: \: \to \: \: \: { \bold{n=total \: \: term}}$

• According to the first condition –

$\\ \implies { \bold{ T_{4} = p }}$

$\\ \implies { \bold{ a {r}^{4 - 1} = p }}$

$\\ \implies { \bold{ a {r}^{3} = p \: \: \: \: \: - - - - eq.(1) }}$

• According to the second condition –

$\\ \implies { \bold{ T_{7} = q }}$

$\\ \implies { \bold{ a {r}^{7 - 1} = q }}$

$\\ \implies { \bold{ a {r}^{6} = q \: \: \: \: \: - - - - eq.(2) }}$

• According to the third condition –

$\\ \implies { \bold{ T_{10} = r }}$

$\\ \implies { \bold{ a {r}^{10 - 1} = r }}$

$\\ \implies { \bold{ a {r}^{9} = r \: \: \: \: \: - - - - eq.(3) }}$

• Now Let's take L.H.S. –

$\\ \: \: \: { \bold{ = {q}^{2} }}$

$\\ \: \: \: { \bold{ = {(a {r}^{6} )}^{2} \: \: \: \: \: \: \: \: \: \: \: [using \: \: eq.(2)]}}$

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

• We should write this as –

$\\ \: \: \: { \bold{ = ( a {r}^{3} )(a {r}^{9}) }}$

$\\ \: \: \: { \bold{ = (p)(r) \: \: \: \: \: \: \: \: \: \: \: [using \: \: eq.(1) \: \: and \: \: eq.(3)]}}$

$\\ \: \: \: { \bold{ = pr}}$

$\\ \: \: \: { \bold{ = R.H.S. \: \: \: \: \: \: \: (Hence \: \: proved)}}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• Fourth term = p
• seventh term = q
• tenth term = r

${\bf{\blue{\underline{To\:Prove:}}}}$

• q² = pr

${\bf{\blue{\underline{Formula\:Used:}}}}$

$\dagger \: \boxed {\sf{ T_{n} = a {r}^{n - 1} }}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ t_{4} = {a} {r}^{4 - 1} }}$

$: \implies\boxed{\sf{ p= {a} {r}^{3} }}$

$: \implies{\sf{ t_{7} = {a} {r}^{7 - 1} }}$

$: \implies\boxed{\sf{ q= {a} {r}^{6} }}$

$: \implies{\sf{ t_{10} = {a} {r}^{10 - 1} }}$

$: \implies\boxed{\sf{ r= {a} {r}^{9} }}$

$\dagger \: \: \underline{\mathfrak{ according \: to \: the \: question}}$

$: \implies{\sf{ {q}^{2} = pr}}$

Put value of p ,r and q in given condition,

$: \implies{\sf{ {( a{r}^{6} })^{2} = a {r}^{3} \times a {r}^{9} }}$

$: \implies{\sf{ { {a}^{2} \times {r}^{6 \times 2} }= a {r}^{3} \times a {r}^{9} }}$

$: \implies{\sf{ { {a}^{2} \times {r}^{12} }= {a}^{2} \times {r}^{12} }}$

$\dagger \: \: \: {\mathfrak{ \purple{hence \: proved}}}$

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$\bigstar \: \: \: \underline{\mathfrak{ \green{ Geomatric \: Progression}}}$

• A series in which the result of dividing any term by previous term(if any) is same ,is called G.P.