# Show that $a_1, a_2, ……. a_n,……$ form an AP where $a_n$ is defined as below:(i) $a_n = 3 + 4n$(ii) $a_n = 9 - 5n$ Also find the sum of the first 15 terms in each case.

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To do:

We have to show that $a_1, a_2, ……. a_n,……$ forms an AP and find the sum of the first 15 terms in each case.

Solution:

(i) Here,

$a_{n}=3+4 n$

This implies,

$a_{1}=3+4 \times 1$

$=3+4$

$=7$

$a=7$

$a_{2}=3+4 \times 2$

$=3+8$

$=11$

$a_{3}=3+4 \times 3$

$=3+12$

$=15$

This implies,

$d=a_{2}-a_{1}$

$=11-7$

$=4$

Therefore,

$a_1, a_2, ……. a_n,……$ forms an AP

Number of terms $=15$

We know that,

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

$S_{15}=\frac{15}{2}[2 a+(15-1) d]$

$=\frac{15}{2}[2 \times 7+(15-1) \times 4]$

$=\frac{15}{2}[14+14 \times 4]$

$=\frac{15}{2}[14+56]$

$=\frac{15}{2} \times 70$

$=15 \times 35$

$=525$

The sum of the first 15 terms is $525$.

(ii) To find the given sequence we have to substitute $n=1, 2, 3.....$ in $a_n=9-5n$.

Therefore,

$a_1=9-5(1)$

$=9-5$

$=4$

$a_2=9-5(2)$

$=9-10$

$=-1$

$a_3=9-5(3)$

$=9-15$

$=-6$

$a_4=9-5(4)$

$=9-20$

$=-11$

The sequence formed is $4, -1, -6, -11,.....$.

For the given sequence to form an A.P., the difference between any two consecutive terms should be equal.

Here,

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

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

$d=a_4-a_3=-11-(-6)=-11+6=-5$

This implies,

$a_2-a_1=a_3-a_2=a_4-a_3=d$

Therefore, the given sequence forms an A.P.

We know that,

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

$S_{15}=\frac{15}{2}[2 a+(15-1) d]$

$=\frac{15}{2}[2 \times 4+(15-1) \times (-5)]$

$=\frac{15}{2}[8+14 \times (-5)]$

$=\frac{15}{2}[8-70]$

$=\frac{15}{2} \times (-62)$

$=15 \times (-31)$

$=-465$

The sum of the first 15 terms is $-465$.

Updated on 10-Oct-2022 13:20:30