Question

If 10th term from the end in the A.P, 5,8,11,..is 95, then the number of terms in the A.P.

Answer 1

Answer:

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Step-by-step explanation:

The nth term of an A.P, tn = a + (n-1) × d

a ➡️ The first term of an A.P

n ➡️ The number of terms in an A.P

d ➡️ The common difference of an A.P

d = t2 - t1

Given:

The 10th term of an A.P is 95

A.P is 5,8,11,......

Solution:

d = 8 - 5

d = 3

a = 5

t10 = 5 + (n - 1) × 3

95 = 5 + 3n - 3

95 = 3n + 2

3n = 93

n = 93÷3

n = 31

The number of terms of this A.P is 31

HOPE U LIKE IT !!✌✌

Answer 2

The value of n (number of terms) will be "31".

Step-by-step explanation:

$The \ given \ series \ is:\\5,8,11,..\\10th \ term,T10=95\\First \ term,a,a1=5\\Second \ term,a2=8\\Difference,d=a2-a1=8-5\\d=3\\Number \ of \ terms,n=?\\As \ we \ know,\\T10=a+(n-1)d\\On \ putting \ the \ values \ in \ the \ above \ formula,we \ get\\95=5+(n-1)3\\95=5+3n-3\\95=2+3n\\On \ subtracting \ "2" \ from \ both \ sides,we \ get\\95-2=2+3n-2\\93=3n\\n=\frac{93}{3} \\n=31$

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