NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3

Question 3

Formulas used for an AP:

nth term: \(a_n = a + (n - 1)d\)

Sum of first n terms: \(S_n = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}(a + l)\), where \(l = a_n\) is the nth term.

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

Use \(a_n = a + (n - 1)d\):

\[50 = 5 + (n - 1)3\]

\[50 - 5 = 3(n - 1) \Rightarrow 45 = 3(n - 1)\]

\[n - 1 = 15 \Rightarrow n = 16\]

Now \(S_n = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{16}{2}[2\cdot 5 + 15\cdot 3]\)

\[S_{16} = 8(10 + 45) = 8 \cdot 55 = 440\]

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

\[a_{13} = a + 12d = 35\]

\[7 + 12d = 35 \Rightarrow 12d = 28 \Rightarrow d = \dfrac{7}{3}\]

Now \(S_{13} = \dfrac{13}{2}[2a + 12d] = \dfrac{13}{2}[14 + 12 \cdot \dfrac{7}{3}]\)

\[14 + 28 = 42 \Rightarrow S_{13} = \dfrac{13}{2} \cdot 42 = 13 \cdot 21 = 273\]

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

\[a_{12} = a + 11d = 37\]

\[a + 11\cdot 3 = 37 \Rightarrow a + 33 = 37 \Rightarrow a = 4\]

\[S_{12} = \dfrac{12}{2}[2a + 11d] = 6[2\cdot 4 + 11\cdot 3] = 6(8 + 33) = 6 \cdot 41 = 246\]

(iv) Given \(a_3 = 15\), \(S_{10} = 125\)

From \(a_3 = a + 2d = 15\), we get \(a = 15 - 2d\).

Also, \(S_{10} = \dfrac{10}{2}[2a + 9d] = 5(2a + 9d) = 125\Rightarrow 2a + 9d = 25\).

Substitute \(a = 15 - 2d\):

\[2(15 - 2d) + 9d = 25 \Rightarrow 30 - 4d + 9d = 25\]

\[30 + 5d = 25 \Rightarrow 5d = -5 \Rightarrow d = -1\]

Then \(a = 15 - 2(-1) = 17\).

10th term: \(a_{10} = a + 9d = 17 + 9(-1) = 8\).

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

\[S_9 = \dfrac{9}{2}[2a + 8d] = 75\]

\[\Rightarrow 9(2a + 40) = 150 \Rightarrow 2a + 40 = \dfrac{150}{9} = \dfrac{50}{3}\]

\[2a = \dfrac{50}{3} - 40 = \dfrac{50}{3} - \dfrac{120}{3} = -\dfrac{70}{3}\Rightarrow a = -\dfrac{35}{3}\]

\[a_9 = a + 8d = -\dfrac{35}{3} + 8\cdot 5 = -\dfrac{35}{3} + \dfrac{120}{3} = \dfrac{85}{3}\]

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

\[S_n = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}[4 + 8(n - 1)]\]

\[= \dfrac{n}{2}(8n - 4) = 2n(2n - 1) = 90\]

So, \[4n^2 - 2n - 90 = 0 \Rightarrow 2n^2 - n - 45 = 0\]

Discriminant: \(1 + 4 \cdot 2 \cdot 45 = 1 + 360 = 361 = 19^2\).

\[n = \dfrac{1 \pm 19}{4}\Rightarrow n = 5\; (\text{positive integer})\]

Then \(a_n = a + (n - 1)d = 2 + 4 \cdot 8 = 34\).

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

From \(S_n = \dfrac{n}{2}(a + l)\):

\[210 = \dfrac{n}{2}(8 + 62) = \dfrac{n}{2} \cdot 70 = 35n\]

\[35n = 210 \Rightarrow n = 6\]

Now \(a_n = a + (n - 1)d = 62\Rightarrow 8 + 5d = 62\Rightarrow 5d = 54\Rightarrow d = \dfrac{54}{5}\).

(viii) Given \(a_n = 4\), \(d = 2\), \(S_n = -14\)

Use \(S_n = \dfrac{n}{2}(a + l)\):

\[-14 = \dfrac{n}{2}(a + 4) \Rightarrow n(a + 4) = -28\]

Also, \(a_n = a + (n - 1)2 = 4\Rightarrow a = 4 - 2(n - 1) = 6 - 2n\).

Then \(a + 4 = 6 - 2n + 4 = 10 - 2n\), so:

\[n(10 - 2n) = -28\Rightarrow 2n(5 - n) = -28\Rightarrow n(5 - n) = -14\]

\[-n^2 + 5n + 14 = 0 \Rightarrow n^2 - 5n - 14 = 0\]

Discriminant: \(25 + 56 = 81 = 9^2\).

\[n = \dfrac{5 \pm 9}{2}\Rightarrow n = 7\; (\text{valid positive})\]

Then \(a = 6 - 2n = 6 - 14 = -8\).

(ix) Given \(a = 3\), \(n = 8\), \(S_n = 192\)

\[S_8 = \dfrac{8}{2}[2a + 7d] = 4(2 \cdot 3 + 7d) = 192\]

\[4(6 + 7d) = 192 \Rightarrow 6 + 7d = 48 \Rightarrow 7d = 42 \Rightarrow d = 6\]

(x) Given \(l = 28\), \(S_n = 144\), \(n = 9\)

Use \(S_n = \dfrac{n}{2}(a + l)\):

\[144 = \dfrac{9}{2}(a + 28)\]

\[288 = 9(a + 28) \Rightarrow a + 28 = 32 \Rightarrow a = 4\]

Summary: Each part uses combinations of the AP nth-term and sum formulas, then solves simple linear or quadratic equations to find the required unknowns.