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CAT 2025 Lesson : Progressions - Special Types II

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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} = ?2×51​+5×81​+8×111​+....+35×381​=?

Solution

Note that  12−15=32×5\dfrac{1}{2} - \dfrac{1}{5} = \dfrac{3}{2 \times 5}21​−51​=2×53​ . Similarly   15−18=35×8\dfrac{1}{5} - \dfrac{1}{8} = \dfrac{3}{5 \times 8}51​−81​=5×83​ and so on.

We shall first express the terms with a numerator of
333.

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}2×51​+5×81​+8×111​+....+35×381​

=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)=31​×(2×53​+5×83​+8×113​+......+35×383​)

=
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)31​×(21​−51​+51​−81​+81​−111​+......+351​−381​)

(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}31​×(21​−381​)=31​×3818​=193​

Answer:
319\dfrac{3}{19}193​

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}n natural numbers = n(n+1)2\dfrac{n(n + 1)}{2}2n(n+1)​

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

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

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

5) Sum of first
n\bm{n}n odd numbers =n2n^2n2

∑i=1nxi\sum\limits_{i = 1}^{n} {x_i}i=1∑n​xi​ is a representation of “sum of all xix_ixi​ values, where iii takes all integer values from 111 till nnn (both inclusive)”.

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

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 291×5+2×6+3×7+.....+25×29 ?

Solution

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

⇒
x=1×(1+4)+2×(2+4)+3×(3+4)+.....+25×(25+29) x = 1 \times (1 + 4) + 2 \times (2 + 4) + 3 \times (3 + 4) + .....+ 25 \times (25+29)x=1×(1+4)+2×(2+4)+3×(3+4)+.....+25×(25+29)

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

As this is just addition, we can separate the summation for
n2n^2n2 and 4n.4n.4n.

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

A constant can be taken common outside the summation.

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

We now apply the formula for sum of squares of
nnn natural numbers and sum of nnn 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}x=625(25+1)(50+1)​+4×225(25+1)​ = 25(25+1)2×(17+4)\dfrac{25 (25 + 1)}{2}\times (17 + 4)225(25+1)​×(17+4)

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

Answer:
682568256825

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 7.7+7.77+7.777+ ....n terms_{n \ terms}n terms​

Solution

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

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

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

=
7n7n7n +79\dfrac{7}{9}97​ (0.9+0.99+0.9990.9 + 0.99 + 0.9990.9+0.99+0.999 + .....n terms_{n \ terms}n terms​)

=
7n7n7n +79\dfrac{7}{9}97​ ((1 – 0.1) + (1 – 0.01) + (1 – 0.001) + .... n terms_{n \ terms}n terms​)

=
7n7n7n +79\dfrac{7}{9}97​ [(1+1+11 + 1 + 1 1+1+1+ .... n terms_{n \ terms}n terms​) – (0.1+0.01+0.0010.1 + 0.01 + 0.0010.1+0.01+0.001 + ... n terms_{n \ terms}n terms​)]

=
7n7n7n +79\dfrac{7}{9}97​ [n – (1101+1102+1103+.....n terms) \left( \dfrac{1}{10^1} + \dfrac{1}{10^{2}} + \dfrac{1}{10^{3}}+ ..... _{n \ terms} \right) (1011​+1021​+1031​+.....n terms​)]

aaa =110\dfrac{1}{10}101​ and r=110r =\dfrac{1}{10}r=101​

=
7n7n7n +79\dfrac{7}{9}97​ [n – 0.1×(1−0.1)n(1−0.1)\dfrac{0.1 \times(1 - 0.1)^n}{(1 - 0.1)}(1−0.1)0.1×(1−0.1)n​ ] = 7n7n7n +79\dfrac{7}{9}97​ n[n – (1−0.1)n9\dfrac{(1 - 0.1)^n}{9}9(1−0.1)n​ ]

=
70n9+7×(1−0.1)n81\dfrac{70n}{9} + \dfrac{7 \times (1 - 0.1)^n}{81}970n​+817×(1−0.1)n​

Answer:
70n9+7×(1−0.1)n81\dfrac{70n}{9} + \dfrac{7 \times ( 1 - 0.1)^n}{81}970n​+817×(1−0.1)n​

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