Question

For an AP t6=-10 and t14=-34 find the value of t10... please help....

Answer 1

Answer:

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Answer 2

Concept Introduction: Arithmetic progression is the progression in which difference between the two numbers is constant

Given:

We have been Given:

$t6 = - 10 \\ t14 = - 34$

To Find:

We have to Find: Find the value of

$t10$

Solution:

According to the problem, we know to find the value of a specific position in AP, is given by the formula,

$tn = a + (n - 1) \times d$

where, a is the first term in an AP series, n is the given specific position in AP series and d is the common difference.

therefore, first putting the value of

$t6 = a + ( 6- 1) \times d = a + 7d$

now,

$t14 = a + ( 14 - 1) \times d = a + 15d$

therefore now subtracting t6 from t14, we get,

$t14 - t6 = a + 13d - a - 5d =- 34 + 10 \\ = 8 d = - 24$

therefore,

$d = - 3$

now, substituting the value of d in any equation above,

$a + 5d = - 10 \\ a + 5\times( - 3) = - 10 \\ a - 15 = - 10 \\ a = 15- 10 \\ a = 5$

Now, after getting all the values find the number in t10

$t10 = 5 + (10 - 1) \times ( - 3) \\ = 5 + 9 \times ( - 3) \\ = 5 - 27 \\ = - 22$

hence,

$t10 = - 27$

Final Answer: The value of t10 is

$t10 = - 22$

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