Example 21: There are \(80 Aps\) \(\left( {{A_1},{\rm{ }}{A_2},{\rm{ }}{A_3},.....{\rm{ }}{A_{80}}} \right)\); such that there first terms are \(\left( {{1^2},{\rm{ }}{2^2},{\rm{ }}{3^2}......{\rm{ }}{{80}^2}} \right)\) respectively and common differences are \(\left( {{1^3},{\rm{ }}{2^3},{\rm{ }}{3^3},{\rm{ }}{4^3},{\rm{ }}...{\rm{ }}{{80}^3}} \right)\) respectively. If each AP has 10 terms, find the sum of all the terms of 80 APs  

Solution: Suppose sum of the different APs are represented by \({S_1},{S_2},{S_3},{\rm{ }} \ldots {S_{80}}\) 

\({S_1}\) = \(\frac{{10}}{2}\left[ {2 \times {1^2} + 9 \times {1^3}} \right] = 5\left[ {2 \times {1^2} + 9 \times {1^3}} \right]\)

\({S_2}\) = \(\frac{{10}}{2}\left[ {2 \times {3^2} + 9 \times {3^3}} \right] = 5\left[ {2 \times {3^2} + 9 \times {3^3}} \right]\)

\({S_3}\) = \(\frac{{10}}{2}\left[ {2 \times {3^2} + 9 \times {3^3}} \right] = 5\left[ {2 \times {3^2} + 9 \times {3^3}} \right]\)

\( \Rightarrow \)S1+ S2+ S3+…SN = \(\sum {\,{S_N} = \frac{{10}}{2}\left[ {2\sum {{N^2} + 9\sum {{N^3}} } } \right]} \)

= \(\frac{{10}}{2}\left[ {\frac{{2 \times 80 \times 81 \times 161}}{6} + 9 \times {{\left( {\frac{{80 \times 81}}{2}} \right)}^2}} \right]\) = 139975.

Example 22:  There are 50 GPs \(\left( {{G_1},{\rm{ }}{G_2},{\rm{ }}{G_3},{\rm{ }}......{\rm{ }}{G_{50}}} \right)\). First term of each \(GP\) are (3, 8, 15, 24, 35, 48, 63, 80.) respectively; and common ratio of each \(GP\) is \(\left( {\frac{1}{{{2^2}}},\frac{1}{{{3^2}}},\frac{1}{{{4^2}}},\frac{1}{{{5^2}}},\frac{1}{{{6^2}}},\frac{1}{{{7^2}}},........} \right)\) respectively. If each \(GP\) is having infinite terms find the sum of all the terms of all the 50 \(GPs\)

Solution:  Suppose sums of the different series are given by 

\({S_1}\) = \(\frac{3}{{1 - \frac{1}{{{2^2}}}}} = 4\), 

\({S_2}\) \(\frac{8}{{1 - \frac{1}{{{3^2}}}}} = 9\),

\({S_3}\) = \(\frac{{15}}{{1 - \frac{1}{{{4^2}}}}} = 16\) 

\( \Rightarrow {S_N} = {(N + 1)^2}\)

Hence \({S_1} + {S_2} + {S_3} + {\rm{ }} \ldots  \ldots ..{S_{50}} = {\rm{ }}{2^2} + {\rm{ }}{3^2} + {\rm{ }}{4^2} + {\rm{ }} \ldots {51^2}\)

= \(\left( {{1^2} + {2^2} + {3^2} + {{....51}^2}} \right) - {1^2}= 45525\)

Example 23: Find the sum of the series:

\(1 \times {3^1} + {\rm{ }}2 \times {3^2} + {\rm{ }}3 \times {3^3} + {\rm{ }}4 \times {3^4} + {\rm{ }}....{\rm{ }} + {\rm{ }}100 \times {3^{100}}\)

Solution:  The question is based on difference method. 

\(S = 1 \times {3^1} + 2 \times {3^2} + 3 \times {3^3} + .....100 \times {3^{100}}\)……....(1) 

\(3S = 1 \times {3^2} + 2 \times {3^3} + .....99 \times {3^{100}} + 100 \times {3^{101}}\)….(2)

Subtracting (1) from (2) 

\(2S = 100 \times {3^{101}} - \left[ {{3^1} + {3^2} + {3^3} + ........... + {3^{100}}} \right]\)

\( \Rightarrow S = \frac{{199}}{4} \times {3^{101}} + \frac{3}{4}\)

Example 24: Find the sum of the series: 

\(1 - 2 - 3 - 4 + 2 - 3 - 4 - 5 + 3 - 4 - 5 -  6 + 4 - 5 - 6 - 7 +  .....  121\) terms.

Solution: In the above series a pattern is observed in a cycle of each four terms. The series can be written as:

\((- 8 - 10 - 12 - …….30\,\ terms) + 31\)

\(= -2(4 + 5 + 6 + …..33) + 31 = - 1079\)

Example 25: How many numbers from \(720\) to 2400 are neither divisible by \(‘6’\) nor by \(‘8’\) nor by \(10\).

Solution: Suppose \(N(k)\) represents the numbers which are multiple of \(k\), then 

\(N(6 \cup 8 \cup 10) = N(6) + N(8) + N(10) - N(24) - N(30) - N(40) + N(120) \) 

\(N(6) = \frac{{2400 - 720}}{6} + 1 = 281\)

Similarly \(N(8) = 211, N (10) = 169, N (24) = 71, N(30) = 57, N(40) = 43, N(120) = 15\)

\(N(6 \cup 8 \cup 10) = 281 + 211 + 169 - 71 - 57 - 43 + 120 = 505\).

Total numbers from 720 to 2400 are 1681.

Number of numbers which are not multiple of 6, 8 or 10 are 1681 - 505 = 1176.

Example 26: Sum of 160 numbers in an \(AP\) is 14400. If the numbers are arranged in ascending order then the sum of last 40 numbers is \(\frac{1}{3}rd\) of the total. Find the smallest of the numbers.

Solution: After arranging the numbers in ascending order, the numbers remain in \(AP\), Given \({S_{160}} = {\rm{ }}14400\)

\({S_{121}}{\,_{to{\rm{ }}160}} = {\rm{ }}4800\), hence \({S_{120}} = {\rm{ }}9600\) 

Assuming first term (smallest term) \(a\) and common difference as \(d\), 

\( \Rightarrow \frac{{160}}{2}\left[ {2a + 159d} \right] = 14400\) ...(1)

\(\frac{{120}}{2}\left[ {2a + 119d} \right] = 9600\) …(2)

Solving (1) and (2) \(a = 50.25\) 

Example 27: If \({A_1} = {\rm{ }}1\) and \({A_{n{\rm{ }} + {\rm{ }}1}} = {\rm{ }}2{A_n} + {\rm{ }}5\), where \(n\) is a natural number, find \({A_{100}}\).

Solution: Putting \(n = 1, 2, 3,\) ….. we get

\({A_2} = {\rm{ }}7{\rm{ }} = {\rm{ }}6 \times {2^1}-{\rm{ }}5,{\rm{ }}{A_3} = \;19{\rm{ }} = {\rm{ }}6 \times {2^2}-{\rm{ }}5\)

\({A_4} = {\rm{ }}43{\rm{ }} = {\rm{ }}6 \times {2^3}-{\rm{ }}5\) and so on.

Thus \({A_{100}}\; = {\rm{ }}6 \times {2^{99}}-{\rm{ }}5\)

Example 28: If the ratio of compound interest for \({199^{th}}\) year to that of \({190^{th}}\) year’s is 512 (interest being compound annually), find rate% of interest.

Solution: Suppose the principal amount is \(P\) and the amount after 190 years and 199 years will be \({A_{190}}\) and \({A_{199}}\) 

\(\)\({A_{190}} = P{\left( {1 + \frac{r}{{100}}} \right)^{190}}\) and \({A_{199}} = P{\left( {1 + \frac{r}{{100}}} \right)^{199}}\)

Thus the compound interests for the year \({199^{th}}\) and \({190^{th}}\) will be: \(\left( {{A_{190}}-{A_{189}}} \right)\) and \(\left( {{A_{199}}-{A_{198}}} \right)\)

\(\frac{{C{I_{199}}}}{{C{I_{190}}}} = \frac{{P{{\left( {1 + \frac{r}{{100}}} \right)}^{199}} - P{{\left( {1 + \frac{r}{{100}}} \right)}^{198}}}}{{P{{\left( {1 + \frac{r}{{100}}} \right)}^{190}} - P{{\left( {1 + \frac{r}{{100}}} \right)}^{189}}}} = {\left( {1 + \frac{r}{{100}}} \right)^9}\)

Given that \({\left( {1 + \frac{r}{{100}}} \right)^9}= 512\) \( \Rightarrow r = {\rm{ }}100\% \)

Example 29: If all the multiples of 6 (natural numbers) are written, one beside the other, in following manner \(N = 61218243036.....\), find the \({6025^{th}}\) digit from beginning for number \(N\)

Solution:  Single digit number of the type \(6k\) is \(1\)

Two digit numbers of the type \(6k\) are \(\frac{{96 - 12}}{6} + 1 = 15\). Number of digits used \( = {\rm{ }}15 \times 2{\rm{ }} = {\rm{ }}30\). 

Three digit numbers of the type \(6k\) are \(\frac{{996\,\, - \,\,102}}{6}\,\, + \,\,1\,\,= 150\). Number of digits used \( = {\rm{ }}150 \times 3{\rm{ }} = {\rm{ }}450\).

Total digits used till now are \(450 + 30 + 1 = 481\).

Remaining digits \(= 6025 - 481 = 5544\) digits. But now each number will have 4 digits, number of such numbers = 1386. Hence, \({6025^{th}}\) digit will be last digit of \({1386^{th}}\) term of series \(A\) where, \(A = 1002, 1008, 1014,……\)

\({T_{1386}} = {\rm{ }}1002{\rm{ }} + {\rm{ }}1385{\rm{ }} \times {\rm{ }}6{\rm{ }} = {\rm{ }}9312\)