If {p^{th}},;{q^{th}},;{r^{th}} and {s^{th}} | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a) If $a$ and $d$ be the first term and common difference of the $A.P.$

Then ${T_p} = a + (p - 1)d,\;$

${T_q} = a + (q - 1)d$

and ${T_r} = a

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(a) If $a$ and $d$ be the first term and common difference of the $A.P.$

Then ${T_p} = a + (p - 1)d,\;$

${T_q} = a + (q - 1)d$

and ${T_r} = a + (r - 1)d$.

If ${T_p},\;{T_q},\;{T_r}$ are in $G.P.$

Then its common ratio $R = \frac{{{T_q}}}{{{T_p}}} = \frac{{{T_r}}}{{{T_q}}} = \frac{{{T_q} - {T_r}}}{{{T_p} - {T_q}}}$

$ = \frac{{[a + (q - 1)d] - [a + (r - 1)d]}}{{[a + (p - 1)d] - [a + (q - 1)d]}} = \frac{{q - r}}{{p - q}}$

Similarly, we can show that $R = \frac{{q - r}}{{p - q}} = \frac{{r - s}}{{q - r}}$

Hence $(p - q),\;(q - r),\;(r - s)$ be in $G.P.$

If ${p^{th}},\;{q^{th}},\;{r^{th}}$ and ${s^{th}}$ terms of an $A.P.$ be in $G.P.$, then $(p - q),\;(q - r),\;(r - s)$ will be in

Similar Questions

If ${a^2} + a{b^2} + 16{c^2} = 2(3ab + 6bc + 4ac)$, where $a,b,c$ are non-zero numbers. Then $a,b,c$ are in

Consider two G.Ps. $2,2^{2}, 2^{3}, \ldots$ and $4,4^{2}, 4^{3}, \ldots$ of $60$ and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}{8}}$, then $\sum_{ k =1}^{ n } k (n- k )$ is equal to.

The ${20^{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + .......$ will be

If the ${10^{th}}$ term of a geometric progression is $9$ and ${4^{th}}$ term is $4$, then its ${7^{th}}$ term is

The sum of a $G.P.$ with common ratio $3$ is $364$, and last term is $243$, then the number of terms is