Question

find the 12th term of the A P with sum of its 12 terms is 75 and 11 terms is 62​

Answer 1

Answer:

Step-by-step exp

12th term is 62

Answer 2

Answer:

Step-by-step explanation:

Consider the first term is $a$ and a common difference is $d$,

Then the $\text{n}^{\text{th}}\text{ term}=a+(n-1)d$

therefore the Sum of Arithmetic series is

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

Sum of its 12 terms is

$S_{12}=\frac{12}{2}[2a+(12-1)d]\\\\\Rightarrow75=6\times[2a+11d]\\\\\Rightarrow75=12a+66d\text{......................equation-1}$

Sum of its 11 terms is

$S_{11}=\frac{11}{2}[2a+(11-1)d]\\\\\Rightarrow62=\frac{11}{2} \times[2a+10d]\\\\\Rightarrow62=\frac{11}{2} \times2[a+5d]\\\\\Rightarrow62=11a+55d\text{......................equation-2}$

Now multiply equation-1 by 11 we get,

$75\times11=11(12a+66d)\\\\825=132a+726d.......................\text{equation-3}$

Multiply equation-2 by 12 and subtract them

$62\times12=12(11a+55d)\\\\744=132a+660d\text{....................equation-4}$

Now subtract equation-4 from equation-3

$(132a+726d)-(132a+660d)=825-744\\\\(132a+726d-132a-660d)=81\\\\66d=81\\\\d=\frac{81}{66}$

Put the value of d in equation -1 we get

$75=12a+66d\\\\75=12a+66\times\frac{81}{66}\\\\75=12a+81\\\\12a=75-81\\\\12a=-6\\\\a=-\frac{1}{2}$

Therefore the 12 term is

$\text{12}^{\text{th}}\text{ term}=-\frac{1}{2}+(12-1)\times\frac{81}{66}=-\frac{1}{2}+\frac{81}{6}=13$