Question

The sum of first three terms of an ap is 45 if the product of first and third term exceeds the second term by 41 find date

Answer 1

Let the first three terms be (a-d), (a), and (a+d).

Given that, the sum of these terms is equal to 45.

$\tt \implies (a - d) + a + (a + d) = 45 \\ \\ \tt \implies a - \cancel{d} + a + a + \cancel{d} = 45 \\ \\ \tt \implies 3a = 45 \\ \\ \tt \implies a = 15$

Now, the second situation given to us is that the product of the first and third term exceeds the second term by 41.

$\tt \implies (a - d)(a + d) = a + 41 \\ \\ \tt \implies {a}^{2} - {d}^{2} = a + 41 \\ \\ \tt \implies {(15)}^{2} - {d}^{2} = 15 + 41 \\ \\ \tt \implies {d}^{2} = 225 - 56 \\ \\ \tt \implies {d}^{2} = 169 \\ \\ \tt \implies d = 13$

So, putting in the values, the AP we get is:

AP => 2, 15, 28, ...

Answer 2

Answer:

AP: 2, 15, 28, ...

Step-by-step explanation:

We have been given that the sum of first terms of an AP is 45.

$S_3$ = 45

Let the first three terms to be a - d, a and a + d where 'd' is the common difference.

According to Question:-

(a - d) + a + (a + d) = 45

a - d + a + a + d = 45

3a = 45

a = $\frac{45}{3}$

a = 15

We also have been given that the product of 1st and 2nd term exceeds 2nd term by 41.

According to Question:-

(a - d)(a + d) = a + 41

Using the identity :-

a² - b² = (a+b)(a-b)

a² - d² = a + 41

Substitute the value of 'a':-

(15)² - d² = 15 + 41

225 - d² = 56

d² = 225 - 56

d² = 169

d = $\sqrt{169}$

d = 13

Substituting the value of 'd' we get the first 3 terms as:-

2, 15 and 28

AP: 2, 15, 28, ...

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