Question

In an A. P. of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565.
the A.P.​

Answer 1

Solution:

Given:

=> A.P consist of 50 terms.

=> Sum of first 10 terms = 210.

=> Sum of last 15 terms = 2565.

To Find:

=> Find A.P

Formula used:

$\sf{\implies S_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

So, We know that sum of first n terms of an A.P is given by,

$\sf{\implies S_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

Put n = 10, we get

$\sf{\implies S_{10}=\dfrac{10}{2}[2a+9d]}$

$\sf{\implies 210=5(2a+9d)}$

$\sf{\implies 3a+9d=42\;\;\;\;.........(1)}$

Sum of last 15 terms is 2565.

$\sf{\implies Sum\;of\;first\;50\;terms-Sum\;of\;first\;35\;terms = 2565}$

$\sf{\implies S_{50}-S_{35}=2565}$

$\sf{\implies \dfrac{50}{2}[2a+(50-1)d]-\dfrac{35}{2}[2a+(35-1)d]=2565}$

$\sf{\implies 25(2a+49d)-\dfrac{35}{2}(2a+34d)=2565}$

$\sf{\implies 5(2a+49d)-\dfrac{7}{2}(2a+34d)= 513}$

$\sf{\implies 3a+126d=513}$

$\sf{\implies a+42d=171\;\;\;\;........(2)}$

Multiply Eq(2) by 2, we get

$\sf{\implies 2a+84d=342\;\;\;\;........(3)}$

Now, Subtracting Eq(1) from Eq(3), we get

$\sf{\implies 84d-9d=342-42}$

$\sf{\implies 75d=300}$

$\sf{\implies d=\dfrac{300}{75}=4}$

Now, Put the value of d in Eq(2), we get

$\sf{\implies a+42d=171}$

$\sf{\implies a+42(4)=171}$

$\sf{\implies a+168=171}$

$\sf{\implies a=3}$

So, A.P is

=> a = 3

=> a + d = 7

=> a + 2d = 11

=> a + 3d = 15

So, required A.P is 3, 7, 11, 15,..........

Answer 2

[FOR WEBSITE VIEWER]

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore A.P=3,7,11,15...}}$

${\bold{\underline{\underline{Step-by-step\:explantion:}}}}$

• In the given question information givem about an A. P. of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565.

• We have to find the A.P.

$\underline \bold{Given : } \\ \implies Number \: of \: terms(n) = 50 \\ \\ \implies S_{10} = 210 \\ \\ \implies S_{15}(From \: last) = 2565 \\ \\ \underline \bold{To \: Find : } \\ \implies A.P = ?$

• According to given question :

$\bold{For \: sum \: of \: 10th \: term : } \\ \implies S_{10} = \frac{10}{2} (2a + (10 - 1) \times d)\\ \\ \implies 210 = 5(2a + 9d) \\ \\ \implies \frac{210}{5} = 2a + 9d \\ \\ \implies 2a + 9d = 42 - - - - - (1) \\ \\ \bold{For \: sum \: of \: last \: 15 \: terms : } \\ \\ \implies S_{15}(from \: end) = \frac{15}{2} (2 \times a_{36} + (15 - 1) \times d) \\ \\ \implies 2565 = \frac{15}{2} (2 \times (a + 35d) + 14d)\\ \\ \implies 2565 = \frac{15}{2} (2a + 70d + 14d) \\ \\ \implies 2565 \times 2 = 15(2a + 84d) \\ \\ \implies \frac{ \cancel{2565} \times 2} { \cancel{15}} = 2(a + 42d) \\ \\ \implies a + 42d = \frac{171 \times \cancel2}{ \cancel2}$$\\ \implies a + 42d = 171 \\ \\ \implies a = 171 - 42d - - - - - (2) \\ \\ \bold{Putting \: value \: of \: a \: in \: (1)} \\ \implies 2(171 - 42d) + 9d = 42 \\ \\ \implies 342 - 84d + 9d = 42 \\ \\ \implies - 75d=42 - 342 \\ \\ \implies \cancel - 75d = \cancel- 300 \\ \\ \implies d = \frac{300}{75} \\ \\ \bold{\implies d = 4} \\ \\ \bold{Putting \: value \: of \: d \: in \:(1)} \\ \implies 2a + 9 \times 4 = 42 \\ \\ \implies 2a + 36 = 42 \\ \\ \implies 2a + 450 = 484 \\ \\ \implies 2a = 42 - 36 \\ \\ \implies a = \frac{6}{2} \\ \\ \bold{\implies a =3} \\ \\ \bold{\therefore A.P = 3,7,11,15....}$