calendarBack
Quant

/

Modern Maths

/

Progressions
ALL MODULES

CAT 2025 Lesson : Progressions - Special Types II

bookmarked

5.4 Splitting Terms


In these questions, the denominator contains the product of adjacent terms of an Arithmetic Progression. These can be solved by splitting each term and expression it as a difference of two terms.

Example 23

12×5+15×8+18×11+....+135×38=?\dfrac{1}{2 \times 5}+\dfrac{1 }{5 \times 8} + \dfrac{1}{8 \times 11} +....+ \dfrac{1}{35 \times 38} = ?

Solution

Note that  12−15=32×5\dfrac{1}{2} - \dfrac{1}{5} = \dfrac{3}{2 \times 5} . Similarly   15−18=35×8\dfrac{1}{5} - \dfrac{1}{8} = \dfrac{3}{5 \times 8} and so on.

We shall first express the terms with a numerator of 33.

12×5+15×8+18×11+....+135×38\dfrac{1}{2 \times 5} + \dfrac{1 }{5 \times 8} + \dfrac{1}{8 \times 11} +....+ \dfrac{1}{35 \times 38}

=13×(32×5+35×8+38×11+......+335×38) = \dfrac{1}{3}\times \left( \dfrac{3}{2 \times 5} + \dfrac{3}{5 \times 8} + \dfrac{3}{8 \times 11} +......+ \dfrac{3}{35 \times 38} \right)

= 13×(12−15+15−18+18−111+......+135−138)\dfrac{1}{3}\times \left( \dfrac{1}{2} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{8} + \dfrac{1}{8} - \dfrac{1}{11} + ......+ \dfrac{1}{35} - \dfrac{1}{38} \right)

(All terms, other than the first and last terms will get cancelled.)

=13×(12−138)=13×1838=319\dfrac{1}{3}\times \left( \dfrac{1}{2} - \dfrac{1}{38}\right) = \dfrac{1}{3}\times\dfrac{18}{38} = \dfrac{3}{19}

Answer: 319\dfrac{3}{19}

5.5 Using sigma (Summation) and separating terms

The following formulae are commonly used to answer questions in CAT and other tests.

1) Sum of first n\bm{n} natural numbers = n(n+1)2\dfrac{n(n + 1)}{2}

2) Sum of squares of first n n natural numbers = (n(n+1)(2n+1))6\dfrac{(n (n + 1)( 2n + 1))}{6}

3) Sum of cubes of first n n natural numbers = [n(n+1)2]2\left[ \dfrac{n ( n + 1)}{2} \right]^2

4) Sum of first n\bm{n} even numbers = n(n+1)n( n + 1)

5) Sum of first n\bm{n} odd numbers =n2n^2

∑i=1nxi\sum\limits_{i = 1}^{n} {x_i} is a representation of “sum of all xix_i values, where ii takes all integer values from 11 till nn (both inclusive)”.

Therefore,∑i=1nxi=x1+x2+x3+.....+xn\sum\limits_{i = 1}^{n} {x_i = x_1 + x_2 + x_3 + .....+ x_n}

Example 24

What is the sum of 1×5+2×6+3×7+.....+25×291 \times 5 + 2 \times 6 + 3 \times 7 + .....+ 25 \times 29 ?

Solution

Let x=1×5+2×6+3×7+.....+25×29 x = 1 \times 5 + 2 \times 6 + 3 \times 7 + .....+ 25 \times 29

⇒ x=1×(1+4)+2×(2+4)+3×(3+4)+.....+25×(25+4) x = 1 \times (1 + 4) + 2 \times (2 + 4) + 3 \times (3 + 4) + .....+ 25 \times (25 + 4)

⇒ x x =∑n=125n(n+4)\sum\limits_{n = 1}^{25} {n(n + 4)} = ∑n=125(n2+4n)\sum\limits_{n = 1}^{25} {(n^2 + 4n)}

As this is just addition, we can separate the summation for n2n^2 and 4n.4n.

⇒ x x =∑n=125(n2)\sum\limits_{n = 1}^{25} {(n^2)} + ∑n=1254n\sum\limits_{n = 1}^{25} {4n}

A constant can be taken common outside the summation.

⇒ x x =∑n=125n2+(4×∑n=125n)\sum\limits_{n = 1}^{25} {n^2}+(4 \times \sum\limits_{n = 1}^{25} {n})

We now apply the formula for sum of squares of nn natural numbers and sum of nn natural numbers.

⇒ x=25(25+1)(50+1)6+4×25(25+1)2 x =\dfrac{25(25 + 1)(50 + 1)}{6} + 4 \times \dfrac{25 (25 + 1)}{2} = 25(25+1)2×(17+4)\dfrac{25 (25 + 1)}{2}\times (17 + 4)

⇒ x=25×13×21=6825 x = 25 \times 13 \times 21 = 6825

Answer: 68256825

5.6 Recurring digits are increasing

We need to express the terms in a GP and then apply the sum to n terms in a GP formula.

Example 25

What is the sum of 7.7+7.77+7.7777.7 + 7.77 + 7.777 + ....n terms_{n \ terms}

Solution

Let x=7.7+7.77+7.777x = 7.7 + 7.77 + 7.777 + .... n terms_{n \ terms}

= (7+7+77 + 7 + 7 + ....nterms_{n terms}) + (0.7+0.77+0.7770.7 + 0.77 + 0.777 + .... n terms_{n \ terms})

= 7n+7(0.1+0.11+0.1117n + 7 (0.1 + 0.11 + 0.111 + ... nterms_{n terms})

= 7n7n +79\dfrac{7}{9} (0.9+0.99+0.9990.9 + 0.99 + 0.999 + .....n terms_{n \ terms})

= 7n7n +79\dfrac{7}{9} ((1 – 0.1) + (1 – 0.01) + (1 – 0.001) + .... n terms_{n \ terms})

= 7n7n +79\dfrac{7}{9} [(1+1+11 + 1 + 1 + .... n terms_{n \ terms}) – (0.1+0.01+0.0010.1 + 0.01 + 0.001 + ... n terms_{n \ terms})]

= 7n7n +79\dfrac{7}{9} [n – (1101+1102+1103+.....n terms) \left( \dfrac{1}{10^1} + \dfrac{1}{10^{2}} + \dfrac{1}{10^{3}}+ ..... _{n \ terms} \right) ]

aa =110\dfrac{1}{10} and r=110r =\dfrac{1}{10}

= 7n7n +79\dfrac{7}{9} [n – 0.1×(1−0.1n)(1−0.1)\dfrac{0.1 \times(1 - 0.1^n)}{(1 - 0.1)} ] = 7n7n +79\dfrac{7}{9} [n – (1−0.1n)9\dfrac{(1 - 0.1^n)}{9} ]

= 70n9+7×(1−0.1n)81\dfrac{70n}{9} + \dfrac{7 \times (1 - 0.1^n)}{81}

Answer: 70n9+7×(1−0.1n)81\dfrac{70n}{9} + \dfrac{7 \times ( 1 - 0.1^n)}{81}

Note: There is an error in the video in the last 2 steps, where the power n has been put outside the bracket. The correct last 2 steps are provided in the text solution provided above.

Loading...Loading Video....