Question

find the number of terms in AP whose A1 is 12, A6 is 8 and sn is 120




pls reply me fast nd give full solution.​

Answer 1

Answer:

Let a and d are first term and

common difference of an A.P

According to the problem given,

a1 = a = 12----( 1 )

a6 = 8

a + 5d = 8 ---( 2 )

Put ( 1 ) in equation ( 2 ) ,

12 + 5d = 8

5d = 8 - 12

5d = -4

d = -4/5

Now ,

a = 12 , d = -4/5,

Let the number of terms = n

Sn = 120

n/2 [ 2a + ( n -1 )d ] = 120

n/2 [ 2×12 + ( n - 1 )( -4/5 ) ] = 120

4n/2 [ 6 + ( n - 1 ) ( -1/5 ) ] = 120

2n [ 30 -n + 1 ]/5 = 120

n ( 31 - n ) = 120 × ( 5/2 )

31n - n² = 300

n² - 31n + 300 = 0

Step-by-step explanation:

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Answer 2

Answer:

after solving this question we will get a quadratic equation as n²-31n+300=0.

As this equation has no roots so it is not possible to find out the number of terms in the given question