In an AP
(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.

(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