Question

the 1st term of GP is 3 and the last term is 243 and the last but one term before it's 81. find the sum of the series​

Answer 1

$\sf \bold{Given\::}$

$\sf \bullet$ $\sf 1^{st}$term of G.P = 3

$\sf \bullet$ Last term = 243

$\sf \bullet$ One term before last term ( $\sf 2^{nd}$ last term) = 81

$\sf \bold{To\:find\::}$ Sum of series

$\sf \bold{Formula\:used\::}$

$\sf \bullet\: T_{n} = a\times r^{n-1}\\\\\sf \bullet \: S_{n} = \dfrac{a(r^{n}-1)}{r-1}$

Here,

$\sf \bullet$ a = $\sf 1^{st}$

$\sf \bullet$ n = Number of term

$\sf \bullet$ r = common ratio

$\sf \bullet$ $\sf T_{n}$ = $\sf n^{th}$ term

$\sf \bullet$ $\sf S_{n}$ = Sum of n terms

$\sf \bold{Formula\:used\::}$

Let , there are n terms present in given series, that is last term is $\sf n^{th}$ term

Therefore,

$\sf T_{n} = a\times r^{n-1}\\\\\sf \implies T_{n} = 3\times r^{n-1}\\\\\sf \implies 243 = 3 \times r^{n-1}\\\\\sf \implies r^{n-1} = \dfrac{243}{3} = 81\:\: ...... equation 1$

Therefore ,

$\sf 2^{nd}$ last term = $\sf (n-1)^{th}$ term

$\sf \implies T_{n-1} = a \times r^{(n-1)-1}\\\\\sf \implies 81 = 3 \times r^{(n-1)-1}\\\\\sf \implies r^{(n-2)} = \dfrac{81}{3} = 27\:\:\:...... equation 2$

Dividing equation 1 by equation 2

$\sf \implies \dfrac{r^{n-1}}{r^{n-2}} = \dfrac{81}{27}\\\\\sf \implies r^{n-1\:-\:n+2} = 3\\\\\sf \implies r = 3$

Putting value of r in equation 1 we get;

$\sf \implies r^{n-1} = 81\\\\\sf \implies 3^{n-1} = 3^{4}\\\\\sf Comparing\: power\:as\: base\: are\: equal\\\sf \implies n-1 = 4\\\sf \implies n = 4+1 = 5$

Sum of series = $\sf S_{n} = \dfrac{a(r^{n}-1)}{r-1}$

Therefore putting value of a,r,n

$\sf \implies S_{5} = \dfrac{3(3^{5}-1)}{3-1}\\\\\sf \implies S_{5} = \dfrac{3\times (243-1)}{2}\\\\\sf \implies S_{5} = \dfrac{3\times (242)}{2}\\\\\sf \implies S_{5} = 3 \times 121\\\\\sf \implies \bold {S_{5} = 363}\\\\\\\sf \bold{ANSWER\::}\:\:\: \bold {S_{5} = 363}$

Answer 2

$\huge {\mathtt {\pink {Given}}}$

In a geometric progression :

• First term = a = 3
• last term = nth term 243
• last but one term = n-1 th term = 81

$\huge {\mathtt {\green{To\:Find}}}$

• Sum of this geometric progression (Sn)

$\huge {\mathtt {\purple{Solution}}}$

$:\implies \tt T_n = a \times r^{n-1}$

$:\implies\tt T_n = 3 \times r^{n-1}$

$:\implies \tt 243 = 3\times r^{n-1}$

$:\implies{ \mathtt {\boxed{\boxed {r^{n-1 }= \dfrac {243}{3} = 81 ... 1}}}}$

$:\implies\tt T_n-1 = a\times r^{(n-1)-1}$

$:\implies \tt 81 = 3 \times r^{(n-1)-1}$

$:\implies {\mathtt{\boxed{\boxed{ r^{(n-2)} = \dfrac {81}{3} = 27 .... 2}}}}$

$:\implies \tt \dfrac {eq 1}{eq 2}$

$:\implies\tt \dfrac {r^{n-1}}{r^{n-2}}$

$:\implies \tt \dfrac {r^{n - 1}}{r-1}= \dfrac {81}{27}$

$:\implies\tt r^{n - 1-n+2}= 3$

$\huge{ \mathtt {\orange {r = 3 }}}$

Now , Substitute the value of r in equation 1

$\tt r^n-1 = 81$

$\tt 3^n-1 = 3^4$

$\tt:\implies n-1 = 4$

$\huge{ \mathtt {\green {n = 5 }}}$

Now sum of the sereis ....

$:\implies \tt S_n =\dfrac {a \times r^n-1}{r-1}$

$:\implies \tt S_5 =\dfrac {3 \times 3^5-1}{3-1}$

$:\implies \tt S_5 =\dfrac {3 \times 243 -1}{2}$

$:\implies \tt S_5 =\dfrac {3 \times 242}{2}$

$:\implies \tt S_5 = 3 \times 121 = 363$

${\sf {\red {Sum \:of \:5\: terms \:of \:this \:geometric\: progression\: is \:363}}}$