Question

the first term of AP is 5 and 100th term is - 292 . find 50th term of this AP​

Answer 1

Given :

• First term of the AP = 5
• 100th term of the AP = -292

To find :

The 50th term of the AP.

Solution :

To find the 50th term of the AP , first we have to find the Common Difference of the AP.

To find the Common Difference (d) :

We know the formula for nth term of an AP i.e,

$\boxed{\bf{t_{n} = a_{1} + (n - 1)d}}$

Where :

• $\bf{t_{n}}$ = nth term of the AP = -292

• $\bf{a_{1}}$ = First term of the AP = 5

• $\bf{n}$ = No. of terms of the AP = 100

• $\bf{d}$ = Common Difference of the AP

Using the formula for nth term and Substituting the value in it, we get :-

$:\implies \bf{t_{n} = a_{1} + (n - 1)d}$

$:\implies \bf{-292 = 5 + (100 - 1)d}$

$:\implies \bf{-292 = 5 + 99d}$

$:\implies \bf{-292 - 5 = 99d}$

$:\implies \bf{-297 = 99d}$

$:\implies \bf{\dfrac{-297}{99} = d}$

$:\implies \bf{-3 = d}$

$\boxed{\therefore \bf{d = (-3)}}$

Hence the Common Difference of the AP is (-3).

Now by using the nth term of the AP and substituting the values in it, we can find the 50th term of the AP.

To find the 50th term of the AP :

We know the formula for nth term of an AP i.e,

$\boxed{\bf{t_{n} = a_{1} + (n - 1)d}}$

Where :

• $\bf{t_{n}}$ = nth term of the AP

• $\bf{a_{1}}$ = First term of the AP = 5

• $\bf{n}$ = No. of terms of the AP = 50

• $\bf{d}$ = Common Difference of the AP = (-3)

Using the formula for nth term and Substituting the value in it, we get :-

$:\implies \bf{t_{n} = a_{1} + (n - 1)d}$

$:\implies \bf{t_{50} = 5 + (50 - 1)(-3)}$

$:\implies \bf{t_{50} = 5 + 49(-3)}$

$:\implies \bf{t_{50} = 5 + (-147)}$

$:\implies \bf{t_{50} = 5 - 147}$

$:\implies \bf{t_{50} = - 142}$

$\boxed{\therefore \bf{t_{50} = - 142}}$

Hence, the 50th term of the AP is -142.

Answer 2

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In the given problem, we are given 1st and 100th term

In the given problem, we are given 1st and 100th termhere

a = 5

$a_{100} = - 292$

now we will find d using formula

$\implies \: a_{n} = a = (n - 1)d$

So,

$\implies a_{100} = a + (100 - 1)d$

-292 = a + ( 100 - 1 )d

To solve for d

To solve for dSubstituting a = 5 we get

-292 = 5 + 99d

-292 - 5 = 99d

$\frac{ - 297}{99} = d$

$\implies \: d = - 3$

Thus

$\implies \: a = 5$

$\implies \: d = - 3$

$\implies \: n = 50$

substituting the above values in the formula

$\implies a_{n} = a + (n - 1)d$

$\implies a_{50} = 5 + (50 - 1)( - 3)$

$\implies a_{50} = 5 - 147$

$\implies a _ {50} = - 142$

Hence

$\implies a_{50} = - 142$