Question

In an AP:

1. given a = 5, d = 3, an = 50, find n and S_n
2. given a = 7, a₁₃ = 35, find d and S₁₃.
3. given a₁₂ = 37, d = 3, find a and S₁₂.
4. given a₃ = 15, S₁₀ = 125, find d and a₁₀.
5. given d = 5, S₉ = 75, find a and a₉.
6. given a = 2, d = 8, S_n = 90, find n and a_n .
7. given a = 8, a_n = 62, S_n = 210, find n and d.
8. given a_n = 4, d = 2, S_n = –14, find n and a.
9. given a = 3, n = 8, S = 192, find d.
10. given l = 28, S = 144, and there are total 9 terms. Find a.

Answer 1

(i) Given that, \(a = 5\), \(d = 3\), \(a_n = 50\)
As \(a_n = a + (n − 1)d\),
\(∴ 50 = 5 + (n − 1)3\)
\(45 = (n − 1)3\)
\(15 = n − 1\)
\(n = 16\)
\(S_n =\frac  n2[a + a_n]\)

\(S_{16} = \frac {16}{2}[5 + 50]\)
\(S_{16} = 8 \times 55\)
\(S_{16} = 440\)

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(ii) Given that, \(a = 7\), \(a_{13} = 35\) 
As, \(a_n = a + (n − 1) d\)
\(∴ a_{13} = a + (13 − 1) d\)
\(35 = 7 + 12 d\)
\(35 − 7 = 12d\)
\(28 = 12d\)
\(d = \frac {28}{12}\)

\(d = \frac 73\)
\(S_n = \frac n2[a + a_n]\)

\(S_{13} =\frac n2[a + a_{13}]\)

\(S_{13}= \frac {13}{2}[7 + 35]\)

\(S_{13} = \frac {13 \times 42}{2}\)
\(S_{13} = 13 \times 21\)
\(S_{13} = 273\)

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(iii) Given that, \(a_{12} = 37\), \(d = 3\)
As \(a_n = a + (n − 1)d\)
\(a_{12}= a + (12 − 1)3\)
\(37 = a + 33\)
\(a = 4\)
\(S_n = \frac n2[a + a_n]\)

\(S_{12} = \frac {12}{2}[4 + 37]\)
\(S_{12}= 6 \times 41\)
\(S_{12} = 246\)

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(iv) Given that, \(a_3 = 15\), \(S_{10} = 125\)
As, \(a_n = a + (n − 1)d\)
\(a_3 = a + (3 − 1)\)
\(15 = a + 2d\)       ……….(i)
\(S_n = \frac {n}{2}[2a + (n-1)d]\)

\(S_{10} = \frac {10}{2}[2a + (10-1)d]\)
\(125 = 5[2a + 9d]\)
\(25 = 2a + 9d\)    ……….(ii)
On multiplying equation (i) by 2, we obtain
\(30 = 2a + 4d\)       ………..(iii)
On subtracting equation (iii) from (ii), we obtain 
\(−5 = 5d\)
\(d = −1\)
From equation (i),
\(15 = a + 2(−1)\)
\(15 = a − 2\)
\(a = 17\)
\(a_{10} = a + (10 − 1)d\)
\(a_{10}= 17 + (9) (−1)\)
\(a_{10} = 17 − 9 = 8\)

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(v) Given that, \(d = 5\), \(S_9 = 75\)
As, \(S_n = \frac {n}{2}[2a + (n-1)d]\)

\(S_9 =\frac 92[2a + (9-1)5]\)

\(75 = \frac 92(2a + 40)\)
\(25 = 3(a + 20)\)
\(25 = 3a + 60\)
\(3a = 25 − 60\)
\(a = -\frac {35}{3}\)
\(a_n = a + (n − 1)d\)
\(a_9 = a + (9 − 1)5\)
\(a_9 = -\frac {35}{3} + 8 \times 5\)

\(a_9 = -\frac {35}{3} + 40\)

\(a_9 = \frac {-35+120}{3}\)

\(a_9 = \frac {85}{3}\)

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(vi) Given that, \(a = 2\), \(d = 8\), \(S_n = 90\)
As, \(S_n = \frac {n}{2}[2a + (n-1)d]\)

\(90 = \frac n2[2  \times 2 + (n-1)8]\)

\(90 = \frac n2[4 + (n-1)8]\)
\(90 = n [2 + (n − 1)4]\)
\(90 = n [2 + 4n − 4]\)
\(90 = n (4n − 2)\)
\(90= 4n^2 − 2n\)
\(4n^2 − 2n − 90 = 0\)
\(4n^2 − 20n + 18n − 90 = 0\)
\(4n (n − 5) + 18 (n − 5) = 0\)
\((n − 5) (4n + 18) = 0\)
Either \(n − 5 = 0\) or \(4n + 18 = 0\)
\(n = 5\) or \(n = -\frac {18}{4} = -\frac {9}{2}\)
However, \(n\) can neither be negative nor fractional.
Therefore, \(n = 5\)
\(a_n = a + (n − 1)d\)
\(a_5 = 2 + (5 − 1)8\)
\(a_5= 2 + 4 \times 8\)
\(a_5= 2 + 32\)
\(a_5 = 34\)

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(vii) Given that, \(a = 8\), \(a_n = 62\), \(S_n = 210\)
\(Sn = \frac {n}{2}[a + a_n]\)

\(210 = \frac n2[8 + 62]\)

\(210 = \frac n2 \times 70\)
\(n = 6\)
\(a_n = a + (n − 1)d\)
\(62 = 8 + (6 − 1)d\)
\(62 − 8 = 5d\)
\(54 = 5d\)
\(d = \frac {54}{5}\)

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(viii) Given that, \(a_n = 4, d = 2, S_n = −14\)
\(a_n = a + (n − 1)d\)
\(4 = a + (n − 1)2\)
\(4 = a + 2n − 2\)
\(a + 2n = 6\)
\(a = 6 − 2n\)       …….(i)
\(S_n = \frac n2[a + a_n]\)

\(-14 = \frac n2[a + 4]\)
\(−28 = n (a + 4)\)
\(−28 = n (6 − 2n + 4)\)       {From equation (i)}
\(−28 = n (− 2n + 10)\)
\(−28 = − 2n^2 + 10n\)
\(2n^2 − 10n − 28 = 0\)
\(n^2 − 5n −14 = 0\)
\(n^2 − 7n + 2n − 14 = 0\)
\(n (n − 7) + 2(n − 7) = 0\)
\((n − 7) (n + 2) = 0\)
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 obtain
\(a = 6 − 2n\)
\(a = 6 − 2 \times 7\)
\(a= 6 − 14\)
\(a= −8\)

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(ix) Given that, \(a = 3\), \(n = 8\), \(S = 192\)
\(S_n = \frac n2[2a + (n-1)d]\)

\(192 = \frac 82[2 \times 3 + (8-1)d]\)
\(192 = 4 [6 + 7d]\)
\(48 = 6 + 7d\)
\(42 = 7d\)
\(d = 6\)

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(x) Given that, \(l = 28\), \(S = 144\) and there are total of \(9\) terms.
\(S_n =\frac  n2(a+l)\)

\(144 = \frac 92(a+28)\)

\(16 = \frac 12(a+28)\)
\(16 × 2 = a + 28\)
\(32 = a + 28\)
\(a = 32 - 28\)
\(a = 4\)