Question

The sum of first four terms of an AP is 56 . The sum of last four terms of same AP is 112 . The first term of the AP is 11 . Find the number of terms in AP .


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Answer 1

HELLO DEAR,

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d] = 4a + (4n – 10) d



GIVEN that:-

sum of first 4term = 56

4a + 6d = 56

=> 2a + 3d = 28

[ a = 11 ]

=> 2(11) +3d = 28

=> 22 + 3d = 28

=> 3d = 28 - 22

=> 3d = 6

=> d = 2

AND,

Also given that:-

sum of Last term = 112

4a + (4n -10)d = 112

=> 4(n -10)(2) = 112 - 44

=> (4n-10) = 68/2

=> 4n = 44

=> n = 11


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Answer 2

Answer:

Thus the number of terms of A.P. is 11

Step-by-step explanation:

Let the A.P. be a,a+d,a+2d,a+3d, a+(n−2)d,a+(n−1)d.

Sum of first four terms =a+(a+d)+(a+2d)+(a+3d)=4a+6d

Sum of last four terms

=[a+(n−4)d]+[a+(n−3)d]+[a+(n−2)d]+[a+(n−1)d]⇒=4a+(4n−10)d

According to the given condition, 4a+6d=56

⇒4(11)+6d=56[Sincea=11(given)]

⇒6d=12⇒d=2

∴4a+(4n−10)d=112

⇒4(11)+(4n−10)2=112

⇒(4n−10)2=68

⇒4n−10=34

⇒4n=44⇒n=11

Thus the number of terms of A.P. is 11.

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