Question

Show that a1,a2..............an forms an ap when an=3+4n and an =9-5n and also find the sum of first fiftheen terms in each case

Answer 1

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Answer 2

1)$a_1,\ a_2,\ .\ .\ .\ .\ a_n$ are in A.P. when $a_n=3+4n$.

The Sum of first fifteen terms is 525.

2)$a_1,\ a_2,\ .\ .\ .\ .\ a_n$ are in A.P. when $a_n=9-5n$.

The Sum of first fifteen terms is -465.

Step-by-step explanation:

For case 1st.

$a_n=3+4n$

For being in A.P. the common difference(d) should be same for each term.

When n=1

$a_n=3+4n\\\\a_1=3+4\times1=3+4=7$

When n=2

$a_2=3+4\times2=3+8=11$

When n=3

$a_3=3+4\times3=3+12=15$

Common difference(d) is calculated by subtracting 1st term from 2nd term.

$d=a_2-a_1=11-7=4$

Again, $d=a_3-a_2=15-11=4$

Here 'd' is same.

So we can say that $a_1,\ a_2,\ .\ .\ .\ .\ a_n$ are in A.P. when $a_n=3+4n$.

We have to find the sum of first fifteen terms.

$S_n=\frac{n}{2}(2a+(n-1)d)$

On putting the values, we get;

$S_{15}=\frac{15}{2}(2\times7+(15-1)4)=\frac{15}{2}(14+14\times4)=\frac{15}{2}(14+56)=\frac{15}{2}\times70=15\times35=525$

Hence The Sum of first fifteen terms is 525.

For case 2nd.

$a_n=9-5n$

For being in A.P. the common difference(d) should be same for each term.

When n=1

$a_n=9-5n\\\\a_1=9-5\times1=9-5=4$

When n=2

$a_2=9-5\times2=9-10=-1$

When n=3

$a_3=9-5\times3=9-15=-6$

$d=a_2-a_1=-1-4=-5$

Again, $d=a_3-a_2=-6-(-1)=-6+1=-5$

Here 'd' is same.

So we can say that $a_1,\ a_2,\ .\ .\ .\ .\ a_n$ are in A.P. when $a_n=9-5n$.

We have to find the sum of first fifteen terms.

$S_n=\frac{n}{2}(2a+(n-1)d)$

On putting the values, we get;

$S_{15}=\frac{15}{2}(2\times4+(15-1)-5)=\frac{15}{2}(8+14\times-5)=\frac{15}{2}(8-70)=\frac{15}{2}\times-62=15\times-31=-465$

Hence The Sum of first fifteen terms is -465.