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(B) Given the $n^{th}$ term of the sequence is $a_{n}=3+4 n$.Calculating the first few terms:$a_{1}=3+4(1)=7$$a_{2}=3+4(2)=11$$a_{3}=3+4(3)=15$$a_{4}=

Vedclass · Accepted Answer

$525$

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(B) Given the $n^{th}$ term of the sequence is $a_{n}=3+4 n$.Calculating the first few terms:$a_{1}=3+4(1)=7$$a_{2}=3+4(2)=11$$a_{3}=3+4(3)=15$$a_{4}=3+4(4)=19$Checking the difference between consecutive terms:$a_{2}-a_{1}=11-7=4$$a_{3}-a_{2}=15-11=4$$a_{4}-a_{3}=19-15=4$Since the difference $a_{k+1}-a_{k}=4$ is constant for all $k$,the sequence forms an Arithmetic Progression $(AP)$ with the first term $a=7$ and common difference $d=4$.The sum of the first $n$ terms of an $AP$ is given by $S_{n}=\frac{n}{2}[2 a+(n-1) d]$.For $n=15$:$S_{15}=\frac{15}{2}[2(7)+(15-1) 4]$$S_{15}=\frac{15}{2}[14+56]$$S_{15}=\frac{15}{2}(70)$$S_{15}=15 	imes 35 = 525$.

Show that $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ form an $AP$ where $a_{n}$ is defined as $a_{n}=3+4 n$. Also,find the sum of the first $15$ terms.

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