Question

Solve the ques 6. ( i ) and ( ii )

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \mathsf{Question \: \: 6 - b}$



Given that the sum of n terms is $\dfrac{5n^{2}}{2}+\dfrac{3n}{2}$




Assuming nth term as 1st term of the AP.


sum of 1st or 1st term itself = $\dfrac{5(1)^{2}}{2}+\dfrac{3(1)}{2}$


sum of 1st term or 1st itself = $\dfrac{8}{2}$


1st term of the AP = 4



Now, assuming nth term as 2nd term of the AP.


sum of first two terms = $\dfrac{5(2)^{2}}{2}+\dfrac{3(2)}{2}$


1st term + 2nd term = $\dfrac{5\times4}{2}+\dfrac{3\times2}{2}$


4 + 2nd term = 10 + 3


4 + 2nd term = 13


2nd term = 9




Then, 1st term = a = 4 and 2nd term = 9



common difference = 2nd term - 1st term


common difference = 9 - 4


common difference = 5




Therefore,


nth term = a + ( n - 1 )d

nth term = 4 + ( n - 1 )5

nth term = 4 + 5n - 5

nth term = 5n - 1




20th term = a + ( 20 - 1 )d

20th term = 4 + 19( 5 )

20th term = 4 + 95

20th term = 99




Therefore, nth term = 5n - 1

20th term = 99




$\: \: \: \: \: \mathsf{Question \: \: 6 - b}$



Given that the sum of n terms is $\dfrac{3n^{2}}{2}+\dfrac{5n}{2}$



Assuming nth term as 1st term of the AP.



sum of 1st or 1st term itself = $\dfrac{3(1)^{2}}{2}+\dfrac{5(1)}{2}$


sum of 1st term or 1st itself = $\dfrac{8}{2}$


1st term of the AP = 4



Now, assuming nth term as 2nd term of the AP.



sum of first two terms = $\dfrac{3(2)^{2}}{2}+\dfrac{5(2)}{2}$


1st term + 2nd term = $\dfrac{3\times4}{2}+\dfrac{5\times2}{2}$


4 + 2nd term = 6 + 5


4 + 2nd term = 11


2nd term = 11 - 4


2nd term = 7




Then, 1st term = a = 4 and 2nd term = 7



common difference = 2nd term - 1st term


common difference = 7 - 4


common difference = 3



Therefore,


nth term = a + ( n - 1 )d

nth term = 4 + ( n - 1 )3

nth term = 4 + 3n - 3

nth term = 3n + 1




25th term = a + ( 25 - 1 )d

25th term = 4 + ( 24 )3

25th term = 4 + 72

25th term = 76




Therefore,


nth term = 3n + 1

25th term = 76
$\:$

Answer 2

6(i)

Given,

$Sum\;of\;n\;terms=\frac{5n^{2}}{2}+\frac{3n}{2}\\\;\\S_n=\frac{5n^{2}+3n}{2}\\\;\\S_n=\frac{n(5n+3)}{2}\\\;\\S_{n-1}=\frac{(n-1)[5(n-1)+3]}{2}\\\;\\S_{n-1}=\frac{(n-1)(5n-5+3)}{2}\\\;\\S_{n-1}=\frac{(n-1)(5n-2)}{2}\\\;\\S_{n-1}=\frac{5n^{2}-2n-5n+2}{2}\\\;\\S_{n-1}=\frac{5n^{2}-7n+2}{2}$

We know, nth term of  AP,

$T_n=S_n-S_{n-1}\\\;\\T_n=\frac{5n^{2}+3n}{2}-\frac{5n^{2}-7n+2}{2}\\\;\\T_n=\frac{10n-2}{2}\\\;\\\text{20th term of AP,}\\\;\\T_{20}=\frac{10\times20-2}{2}\\\;\\T_{20}=\frac{200-2}{2}\\\;\\T_{20}=\frac{198}{2}\\\;\\T_{20}=99$

6(ii)

Given,

$Sum\;of\;n\;terms=\frac{3n^{2}}{2}+\frac{5n}{2}\\\;\\S_n=\frac{3n^{2}+5n}{2}\\\;\\S_n=\frac{n(3n+5)}{2}\\\;\\S_{n-1}=\frac{(n-1)[3(n-1)+5]}{2}\\\;\\S_{n-1}=\frac{(n-1)(3n-3+5)}{2}\\\;\\S_{n-1}=\frac{(n-1)(3n+2)}{2}\\\;\\S_{n-1}=\frac{3n^{2}+2n-3n-2}{2}\\\;\\S_{n-1}=\frac{3n^{2}-n-2}{2}$

We know, nth term of  AP,

$T_n=S_n-S_{n-1}\\\;\\T_n=\frac{3n^{2}+5n}{2}-\frac{3n^{2}-n-2}{2}\\\;\\T_n=\frac{6n+2}{2}\\\;\\\text{25th term of AP,}\\\;\\T_{25}=\frac{6\times25+2}{2}\\\;\\T_{25}=\frac{150+2}{2}\\\;\\T_{25}=\frac{152}{2}\\\;\\T_{25}=76$