Show that a_{1}, a_{2}, ldots, a_{n}, ldots f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $a_{n}=9-5n$.Calculating the first few terms:$a_{1}=9-5(1)=4$$a_{2}=9-5(2)=-1$$a_{3}=9-5(3)=-6$Checking the common difference:$a_{2}-a_{1}=-

Vedclass · Accepted Answer

$-465$

Vedclass · Answer

(C) Given $a_{n}=9-5n$.Calculating the first few terms:$a_{1}=9-5(1)=4$$a_{2}=9-5(2)=-1$$a_{3}=9-5(3)=-6$Checking the common difference:$a_{2}-a_{1}=-1-4=-5$$a_{3}-a_{2}=-6-(-1)=-5$Since the difference $a_{k+1}-a_{k}$ is constant $(-5)$,the sequence forms an $AP$ with first term $a=4$ and common difference $d=-5$.The sum of the first $n$ terms is given by $S_{n}=\frac{n}{2}[2a+(n-1)d]$.For $n=15$:$S_{15}=\frac{15}{2}[2(4)+(15-1)(-5)]$$S_{15}=\frac{15}{2}[8+14(-5)]$$S_{15}=\frac{15}{2}[8-70]$$S_{15}=\frac{15}{2}(-62)$$S_{15}=15(-31)=-465$.

Show that $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ form an $AP$ where $a_{n}$ is defined as $a_{n}=9-5n$. Find the sum of the first $15$ terms.

Similar Questions

Determine the $AP$ whose $3^{rd}$ term is $5$ and the $7^{th}$ term is $9$.

Find the number of terms in the following $AP$: $18, 15 \frac{1}{2}, 13, \ldots, -47$.

Find the $11^{th}$ term from the last term (towards the first term) of the $AP : 10, 7, 4, \ldots, -62$.

Find the sum of the first $15$ multiples of $8.$

Are $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$ in an $AP$? If they form an $AP$,find the common difference $d$ and write three more terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module