Question

Find the 21st term of an A.P whose 15th term is 25 and 29th term is 46.Show that 29 does not belong to A.P.

Answer 1

Answer: 21st term is 34

Step-by-step explanation:

I have attached the solution (note: An is referred as Tn)

Hope it helps

Please mark brainliest

Regards

Answer 2

Given:

$t_{15} = 25 \\\\ t_{29} = 46$

To Find:

$21_{st}$ term of the same AP

Answer:

Explanation:

We know that,

General term of an AP = a + (n-1)d

In the question , $15_{th}$ and $29_{th}$ terms of AP are given.

$15_{th} \: term \: = a + (n - 1)d \\ 25 = a + (15 - 1)d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ 25 = a + 14d \: \: \: \: \: ...... (1) \: \: \: \: \: \: \:$

Similarly,

$29_{th} \: term \: = a + (n - 1)d \\ 46 = a + (29 - 1)d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ 46 = a + 28d \: \: \: \: \: \: .......(2) \: \: \: \: \:$

Subtracting equation (2) from (1) , we get:

a + 14d - a - 28d = 25 - 46

-14d = -21

$d = \dfrac{21}{14} \\ d = \dfrac{3}{2} \: \:$

Substituting the value of d in equation (1), we get:

$25 = a + 14 \times \dfrac{3}{2} \\ 25 = a + 21 \: \: \: \: \: \: \: \: \: \: \\ a = 4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

To Calculate:

$t_{21} = a + 20d \: \: \: \: \: \: \\ t_{21} = 4 + 20 \times \dfrac{3}{2} \\ t_{21} = 4 + 30 \: \: \: \: \: \: \: \: \: \\ t_{21} = 34 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore, the answer is 34.

Also, we need to prove that 29 is not a term of the above AP.

We know that,

General Term of an AP = a + (n-1)d

where,

a = first term (Can be Integer , fraction or decimal )

n = Number of terms of AP( Can only be natural number i.e. above 1)

d = common difference (Can be Integer , fraction or decimal )

29 = 4 + (n-1)$\dfrac{3}{2}$

29 - 4 = $\dfrac{3}{2}$(n-1)

Cross multiplying, we get:

50 = 3n - 3

3n = 53

n = $\dfrac{53}{3}$

Since , the position of a number of an AP can only be natural number and not fraction.

So, 29 does not belong to the above AP.

Hence Proved.

Other AP Formulas:

nth term of an AP formulas

$\\\\\sf1) \: n_{th} \: term \: of \: any \: AP \: = a + (n - 1)d$

$\sf2)\: n_{th} \: term \: from \: the \: end \: of \: an \: AP \: = a + (m - n)d$

$\sf3)\:n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf4)\:Difference \: of \: two \: terms = (m - n)d$

where m and n is the position of the term in AP

$\sf5)\:Middle\:Term\:of\:an\:AP$

$\sf(i) \: If \: n \: is \: odd = \frac{n + 1}{2}\: th\: term$

$\sf(ii) \:If \: n \: is \: even = \frac{n}{2} \:th \: term \: and \: ( \frac{n}{2} + 1)th \: term$

Sum Formulas

$\\\\\sf1)\:Sum \: of \: first \: n \: terms \: of \: an \:AP = \frac{n}{2} [ \: 2a + (n - 1)d \: ]$

$\sf2)\:Sum \: of \: first \: n \: natural \: numbers = \frac{n(n + 1)}{2}$

$\sf3)\:Sum \: of \: AP \: having \: last \: term = \frac{n}{2} [ \: a + l \: ]$