Question

Hello everyone~














can someone please give me four examples (own questions and answers to those questions) in which the (An) formula is used? (a.k.a an=a+(n-1)d) please give me some questions with answers in which this formula is used. ​

Answer 1

Question: If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.

Solution:

It is given that nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9,. It means Tn = Pn.

By Given Condition, We have:

Tn = a + ( n -1)d

Tp = a + ( n -1)d

⇒ Tn = Tn

⇒ a + ( n -1)d = a + ( n-1)d

⇒ 9 + ( n - 1)-2= 15 + ( n -1)-3

⇒ 9 - 2n + 2 = 15 - 3n + 3

⇒ 11- 2n = 18 - 3n

⇒ 11 - 18 = -3n + 2n

⇒ - 7 = - n

⇒ n = 7

•°• Therefore, Required Value of n is 7.

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Question: How many terms of the AP 20, 19 1/ 3, 18 2 / 3 must be taken so that their sum is 300 and Explain the double answer.

Solution:

Given arithmetic sequence or progressions ;-

◀ Ap :- 20 , 58 / 3 , 56 / 3 ---- 300

Here,

◀ Given;-

first term=20

common difference= 56 / 3 - 20 = - 2 / 3

◀ We know that formula of summation;-

Sn = n / 2 ( 2a + (n-1)d )

◀ 300 = n / 2 [ 2(20) + (n-1)(- 2/3) ]

◀ 600 = n ( 40 - 2n / 3 + 2/3 )

◀ 600 x 3 = n ( 120 - 2n + 2 )

◀ 1800 = n ( 122 - 2n )

◀ 1800=122n-2n²

Arrange in Quadratic form;-

◀ 2n² - 122n + 1800 = 0

Taken common 2 from Quadratic form;-

→ n² - 61n + 900 = 0

→n² - 36n - 25n + 900 = 0

→ n ( n - 36 ) - 25 ( n - 36 ) = 0

→ ( n - 25 ) ( n - 36 ) = 0

→ n = 25 or 36 .

Hence, Two Sum of number obtained = 300

→ Solving both Equaton;-

→ S25 = 25 / 2 ( 40 + (25 - 1) ( - 2 / 3 ) = 25 / 2 ( 40 - 16 )

→ 25 / 2 ( 24 )

→ 25 x 12 = 300

→ S36 = 36 / 2 ( 40 + (36 - 1) ( - 2 / 3 )

→ 18 ( 40 + 35 ( - 2 / 3 )

→ 18 ( 40 - 70 / 3 )

→18 x 50 / 3

→ 6 x 50

→ 300.

‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎Hence, Required numbers=25 and 36

Answer 2

Question1.

The sum of the 4th term and 8th term of an AP is 24 and sum of 6th term and 10th term is 44. Find an AP.

Answer:

$a_{n}$ = a + (n - 1)d

A.T.Q.

=> $a_{4}$ + $a_{8}$ = 24

=> a + (4 - 1)d + a + (8 - 1)d = 24

=> a + 3d + a + 7d = 24

=> 2a + 10d = 24

=> a + 5d = 12

=> a = 12 - 5d ________ (eq 1)

=> $a_{6}$ + $a_{10}$ = 44

=> a + (6 - 1)d + a + (10 - 1)d = 44

=> a + 5d + a + 9d = 44

=> 2a + 14d = 44

=> a + 7d = 22

=> 12 - 5d + 7d = 22 [From (eq 1)]

=> 2d = 10

=> d = 5

Put value of d in (eq 1)

=> a = 12 - 5(5)

=> a = 12 - 25

=> a = - 13

A.P. :

• a = - 13

• a + d = - 13 + 5 = - 8

• a + 2d = - 13 + 2(5) = - 13 + 10 = - 3

A.P : - 13, - 8, - 3 ...

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Question2.

How many 2 digits number are divisible by 7?

Answer:

AP : 14, 21, 28, ...... , 98

a = 14, d = 21 - 14 = 7, $a_{n}$ = 98

$a_{n}$ = a + (n - 1)d

98 = 14 + (n - 1)7

84 = (n - 1)7

12 = n - 1

n = 13

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Question3.

Find AP whose 4th term is 6 and 8th term is 10.

Answer:

$a_{4}$ = 6

a + (4 - 1)d = 6

a + 3d = 6

a = 6 - 3d ______ (eq 1)

$a_{8}$ = 10

a + (8 - 1)d = 10

a + 7d = 10

6 - 3d + 7d = 10 [From (eq 1)]

4d = 4

d = 1

Put value of d in (eq 1)

a = 6 - 3(1)

a = 6 - 3

a = 3

AP : 3, 4, 5 ...

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