Question

How many terms are there in the following Arithmetic Progressions?
(i) 7, 13, 19........205.
(ii) 38, 35, 32.........2.
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Answer 1

AnswEr:-

Your answers are as follows:-

i) 34.

ii) 13.

ExplanaTion:-

Given APs:-

• 7, 13, 19.....205.

• 38, 35, 32......2.

To Find:-

• Number of terms in each of the following.

Formula Used:-

$\\ \tt\longrightarrow a_n = a+(n-1)d.\\ \\ \tt\longrightarrow d = a_2 - a.$

Where,

• $\tt a_n$ = $\tt n^{th}$ term of the AP.

• a = First term of the AP.

• $\tt a_2$ = Second term of the AP.

• d = Common difference of the AP.

• n = Number of terms in the AP.

So Here,

In (i),

• $\tt a_n$ = 205.

• a = 7.

• $\tt a_2$ = 13.

• d = 13-7 = 6.

And we have to find the value of n.

So lets put the above values in the formula:-

$\\ \\ \tt\mapsto a_n = a+(n-1)d .\\ \\ \tt\mapsto 205 = 7+(n-1)6. \\ \\ \tt\mapsto 7+6n-6 = 205. \\ \\ \tt\mapsto 1+6n = 205. \\ \\ \tt\mapsto 6n = 205-1. \\ \\ \tt\mapsto 6n = 204. \\ \\ \tt\mapsto n = \dfrac{\cancel{204}}{\cancel{6}}. \\ \\ \tt\mapsto n=34.$

$\rm\therefore$ There are 34 terms in the AP.

Similarly in(ii),

• $\tt a_n$ = 2.

• a = 38.

• $\tt a_2$ = 35.

• d = 35-38 = -3.

And we have to find the value of n.

So lets put the above values in the formula:-

$\\ \\ \tt\mapsto a_n = a+(n-1)d .\\ \\ \tt\mapsto 2 = 38+(n-1)(-3). \\ \\ \tt\mapsto 38-3n+3 = 2. \\ \\ \tt\mapsto 41-3n = 2. \\ \\ \tt\mapsto -3n = 2-41. \\ \\ \tt\mapsto -3n = -39. \\ \\ \tt\mapsto n = \dfrac{\cancel{-39}}{\cancel{-3}}. \\ \\ \tt\mapsto n=13.$

$\rm\therefore$ There are 13 terms in the AP.