Question

please solve this
need it fast guys​

Answer 1

1)

$t1 = 2. {3}^{1 - 1} + 1 = 2 \times {3}^{0} + 1 \\ t2 = 2. {3}^{2 - 1} + 1 = 2 \times {3}^{1} + 1 \\ t3 = 2. {3}^{3 - 1} + 1 = 2 \times {3}^{2} + 1 \\ . \: \: \: \: \: \: \: \: \: . \: \: \: \: \: \: \: .\: \: \: \: \: \: \: \: \: . \\ . \: \: \: \: \: \: \: \: \: . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .\\ tr = 2 \times {3}^{r - 1} + 1 = 2 \times {3}^{r - 1} + 1 \\ \\ t1 + t2 + t3 + ... + tr = 2 \times \frac{ {3}^{0}( {3}^{r} - 1)}{3 - 1} + r \\ sum \: upto \: r \: terms = {3}^{r} - 1 + r$

8)Let the first term of that GP be A

Then,

$a = {A}^{p - 1} \\ b = {A}^{q - 1} \\ c = {A}^{r - 1} \\ putting \: this \: value \: of \: a \: b \: and \: c \: in \: LHS \: we \: get \\( { {A}^{p - 1} })^{q - r}.( { {A}^{q - 1} })^{r - p}. ({ {A}^{r - 1} })^{p - q} \\ adding \: all \: the \: powers \: we \: get \\ {A}^{0} = 1 = rhs$