Question

Answer any nine out of following 12 questions with calculations:
19. nos. 25 to 36 - 2 marls each)
Given that HCF (306, 657) = 9, find LCM (306, 657).
- Find the HCF of 105 and 1515 by the prime factorisation method. From that, fin
their LCM
and the produe
4
. Find a quadratic polynomial for which the sum of zeroes is
of zeroes is a
3. Which term of the AP 21, 18, 15, ... is 0?
9. Find the sum of first 20 terms of the AP 5, 13, 21,​

Answer 1

Step-by-step explanation:

1) using the formula: HCF X LCM = PRODUCT OF THE NUMBERS

now, substitute:

take LCM as variable 'a' for easier calculation

=> 9 X a = 306 X 657

=>  9 X a = 201042

=>  a = 201042 / 9

=> a =  22338

therefore, LCM = 22338

2) prime factorization can be done by factor tree, long division etc. methods

over here I am using long division method

HCF of 105 & 1515:

=>    105 ) 1515( 14

               -1470

            ---------------

                     35 ) 105 ( 3

                           -105

                           ------------------

                                 0                                                          

                          --------------------

therefore, HCF = 35

using formula,

HCF X LCM = PRODUCT OF NUMBERS

take LCM as variable 'a' for easier calculation

now, substitute:

=> 35 X a = 105 X 1515

=> 35 X a = 159075

=> a = 159075/ 35

=> a = 4545

therefore,

LCM = 4545

product = 35 X 4545 = 159075

3) cannot answer

4)  formula:

$a_n = a+(n-1)d$

AP : 21 , 18 , 15....

=> a = 21 (given)

=> d = (18-21) = (-3)

=>  (15-18) = (-3)

=>  $a_n$ = 0 (given)

solution:

substituting values:

=> 0 = 21 + (n-1) (-3)

=> 0 = 21 + (-3n) (+3)

=> 0= 24 (-3n)

=> -24 = -3n

=> -24/-3 = n

=> 8 = n

therefore, the 8th term of AP : 21 , 18 , 15 = 0

5) AP : 5 , 13 , 21

formula : $s_n = \frac{n}{2} [2a+(n-1)d]$

=> a = 5 (given)

=> n = 20 (given)

=> d = (13-5) = 8

         (21-13) = 8

substituting values

=>$s_n = \frac{20}{2}$ [ 2 X 5 + ( 20-1 )8]

=>$s_n$ = 10 [10 + (19)8]

=> $s_n$ = 10 [ 10 + 152 ]

=> $s_n$ = 10 [ 162 ]

=> $s_n$ = 1620

therefore, the sum of first 20 terms of the AP 5 , 13 , 21  is 1620

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hope this helps :)