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(i)Given a = 5, d = 3, an = 50, find n and \(S_n\).

(ii) Given a = 7, \(a_{13}\) = 35, find d and \(S_{13}\).

(iii) Given \(a_{12}\) = 37, d = 3, find a and \(S_{12}\).

(iv) Given \(a_3\) = 15, \(S_{10}\) = 125, find d and \(a_{10}\).

(v) Given d = 5, \(S_9\)= 75, find a and \(a_9\).

(vi) Given a = 2, d = 8, \(S_n\) = 90, find n and an.

(vii) Given a = 8, an = 62, \(S_n\) = 210, find n and d.

(viii) Given \(a_n\) = 4, d = 2, \(S_n\) = - 14, find n and a.

(ix) Given a = 3, n = 8, S = 192, find d.

(x)Given l = 28, S = 144 and there are total 9 terms. Find a.

Answer :

(i) Given that, a = 5, d = 3, \(a_n\) = 50

As we know, from the formula of the nth term in an AP,

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow \) 50 = 5+(n -1)×3

\(\Rightarrow \) 3(n -1) = 45

\(\Rightarrow \) n -1 = 15

\(\Rightarrow \) n = 16

Now, sum of n terms,

\(\Rightarrow S_n = (a + a_n)\)

\(\Rightarrow S_n = {{16} \over{2}} (5 + 50) = 440\)

(ii) Given that, a = 7, \(a_{13}\) = 35

As we know, from the formula of the nth term in an AP

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow \) 35 = 7+(13-1)d

\(\Rightarrow \) 12d = 28

\(\Rightarrow \) d = 28/12 = 2.33

Now, sum of n terms,

\(\Rightarrow S_n = (a + a_n)\)

\(\Rightarrow S_n = {{13} \over {2}} (7 + 35) = 273\)

(iii)Given that, \(a_{12}\) = 37, d = 3

As we know, from the formula of the nth term in an AP,

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow a_{12}\) = a + (12 - 1)3

\(\Rightarrow \) 37 = a + 33

\(\Rightarrow \) a = 4

Now, sum of nth term,

\(\Rightarrow S_n = (a + a_n)\)

\(\Rightarrow S_n = {{12} \over {2}} (4 + 37)\)

= 246

(iv) Given that, \(a_3\) = 15, \(S_{10}\) = 12

As we know, from the formula of the nth term in an AP,

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow a_3\) = a+(3 - 1)d

\(\Rightarrow\) 15 = a + 2d ………………………….. (i)

Sum of the nth term,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\(\Rightarrow S_{10} = {{10} \over {2}} [2a+(10-1)d] \)

\(\Rightarrow \) 125 = 5(2a+9d)

\(\Rightarrow \) 25 = 2a+9d ……………………….. (ii)

On multiplying equation (i) by 2, we will get;

30 = 2a+4d ………………………………. (iii)

By subtracting equation (iii) from (ii), we get,

-5 = 5d

d = -1

From equation (i)

15 = a + 2(-1)

15 = a - 2

a = 17 = First term

\(\Rightarrow a_{10}\) = a + (10-1)d

\(\Rightarrow a_{10}\) = 17 + (9)(-1)

\(\Rightarrow a_{10}\) = 17 -9 = 8

(v) Given that, d = 5, \(S_9\) = 75

As, sum of n terms in AP is,

\( S_n = {{n} \over {2}} [2a+(n-1)d]\)

Therefore, the sum of first nine terms are;

\(\Rightarrow S_9 = {{9} \over {2}} [2a+(9 -1)5]\)

\(\Rightarrow \) =25 = 3(a+20)

\(\Rightarrow \) 25 = 3a+60

\(\Rightarrow \) 3a = 25-60

\(\Rightarrow \) a = \({{-35} \over {3}}\)

As we know, the nth term can be written as;

\(a_n = a + (n – 1)d\)

=\(a_9 = a + (9 - 1)(5)\)

= \({{-35} \over {3}}\) +8(5)

= \({{-35} \over {3}}\)+40

= (\({{-35 + 120} \over {3}}\))

= \({{85} \over {3}}\)

(vi) Given that, a = 2, d = 8, \(S_n\) = 90

As, sum of n terms in an AP is,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\(\Rightarrow 90 = {{n} \over {2}} [2a +(n -1)d]\)

\(\Rightarrow 180 = n(4 +8n - 8)\)

\(\Rightarrow 180 = n(8n-4) = 8n^2-4n\)

\(\Rightarrow 8n^2 - 4n –180 = 0\)

\(\Rightarrow 2n^2 - n - 45 = 0\)

\(\Rightarrow 2n^2 -10n + 9n - 45 = 0\)

\(\Rightarrow 2n(n - 5) + 9(n - 5) = 0\)

\(\Rightarrow(2n - 9)(2n + 9) = 0\)

So, n = 5 (as it is positive integer)

\(a_5\) = 8 + 5×4 = 34

(vii) Given that, a = 8, \(a_n\) = 62, \(S_n\) = 210

As, sum of n terms in an AP is,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\(\Rightarrow 210 = {{n} \over {2}} (8 +62)\)

\(\Rightarrow \) 35n = 210

\(\Rightarrow \) n = \(\frac{210}{35}\) = 6

Now, 62 = 8 + 5d

\(\Rightarrow \) 5d = 62 - 8 = 54

\(\Rightarrow \) d = \({{54} \over {5}}\) = 10.8

(viii) Given that, nth term, an = 4,

common difference, d = 2, sum of n terms, \(S_n\) = -14.

As we know, from the formula of the nth term in an AP,

\( a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow \) 4 = a+(n -1)2

\(\Rightarrow \) 4 = a+2n-2

\(\Rightarrow \) a+2n = 6

\(\Rightarrow \) a = 6 - 2n ……………………. (i)

As we know, the sum of n terms is;

\(\Rightarrow S_n = \frac{n}{2} (a + a_n) \)

\(\Rightarrow -14 = \frac{n}{2}(a+4)\)

\(\Rightarrow-28 = n(a + 4)\)

\(\Rightarrow-28 = n(6 -2n + 4) \){From equation (i)}

\(\Rightarrow-28 = n(-2n +10)\)

\(\Rightarrow-28 = - 2n^2 + 10n\)

\(\Rightarrow 2n^2 -10n - 28 = 0\)

\(\Rightarrow n^2 -5n -14 = 0\)

\(\Rightarrow n^2 -7n + 2n -14 = 0\)

\(\Rightarrow n(n - 7) + 2(n - 7) = 0\)

\(\Rightarrow (n -7)(n +2) = 0\)

Either n - 7 = 0 or n + 2 = 0

n = 7 or n = -2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we get

\(\Rightarrow \) a = 6 - 2n

\(\Rightarrow \) a = 6 - 2(7)

\(\Rightarrow\) = 6 - 14

Thus, a = -8

(ix) Given that, first term, a = 3,

Number of terms, n = 8

And sum of n terms, S = 192

As we know,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\( \Rightarrow 192 = {{8} \over {2}} [2×3+(8 -1)d]\)

\(\Rightarrow \) 192 = 4[6 + 7d]

\(\Rightarrow \) 48 = 6 + 7d

\(\Rightarrow \) 42 = 7d

\(\Rightarrow \) d = 6

(x) Given that, l = 28,S = 144 and there are total of 9 terms.

Sum of n terms formula,

\(\Rightarrow S_n = {{n} \over {2}} (a + l)\)

\(\Rightarrow 144 = {{9} \over {2}}(a+28)\)

\(\Rightarrow \) (16)×(2) = a + 28

\(\Rightarrow \) 32 = a + 28

\(\Rightarrow \) a = 4

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