Question

In an AP of 50 terms the sum of first 10term is 210and the sum of its last 15term is 2565. Find The AP

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

First term be a

Common diference be d

${\boxed{\sf\:{Arithemetic\;Progressions\;are :- }}}$

a , a + d , a + 2d, a + 48d , a + 49d

$\Large{\boxed{\sf\:{Using\;Formula\;we\;get:-}}}$

${\boxed{\sf\:{Sum\;of\;nth\;terms}}}$

$\tt{\rightarrow\dfrac{n}{2}\times (a+l)}$

$\tt{\rightarrow S_{10}=\dfrac{10}{2}\times (a+a+9d)}$

5(2a + 9d) = 210

2a + 9d = 42 ............... (1)

${\boxed{\sf\:{Last\;15\;term\;from\;36\;to\;50}}}$

$\tt{\rightarrow 36^{th}term=a+35d}$

$\tt{\rightarrow 50^{th}term=a+49d}$

Now,

$\tt{\rightarrow Sum\;of\;last\;15\;term=\dfrac{15}{2}\times (a+35d+a+9d)}$

$\tt{\rightarrow\dfrac{15}{2}\times (2a+84d)=2565}$

2a + 84d = 342 ............... (2)

Now,

${\boxed{\sf\:{Subtracting\;(2)\;from\;(1)\;we\;get,}}}$

75d = 300

$\tt{\rightarrow d=\dfrac{300}{75}}$

d = 4

2a + 9d = 42

2a + 36 = 42

2a = 6

a = 3

Here we get first term and common difference

a = 3

a + d = 3 + 4 = 7

a + 2d = 3 + 2 × 4 = 11

a + 48d = 3 + 48 × 4 = 195

a + 49d = 3 + 49 × 4 = 199

$\boxed{\begin{minipage}{11 cm} Hence\;we\;have \\ \\ \ a=3 \\ \\ a+d=3+4=7 \\ \\ a+2d=3+2\times 4=7 \\ \\ a+48d = 3+48\times 4=195 \\ \\ a+49d=3+49\times 4=199 \end{minipage}}$

Answer 2

Question :--- In an AP of 50 terms the sum of first 10term is 210and the sum of its last 15term is 2565. Find The AP ?

❂ Concept and Formula used :---

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

____________________________

Solution :----

Let First term of AP = a, and common differnce = d ,

→ sum of First 10 terms = 210

→ (10/2)[2a + (10-1)d] = 210

→ 5[2a + 9d] = 210

→ 2a + 9d = 42 ----------------------- Equation (1)

Now, Sum of last 15 terms is given 2565 .

→ in last 15 terms First term = (50-15) + 1 = 36th term .

→ T(36) = a + (36-1)d = a + 35d .

So,

→ sum of last 15 terms = 2565

→ (15/2) [ 2(a+35d) + (15-1)d] = 2565

→ [ 2a + 70d + 14d ] = 2565*2/15

→ 2a + 84d = 342 ----------------------- Equation (2)

______________________________

Subtracting Equation (1) From Equation (2) now, we get,

→ 75d = 300

→ d = 300/75

→ d = 4

Putting value of d in Equation (1) now, we get,

→ 2a + 9*4 = 42

→ 2a = 42-36

→ 2a = 6

→ a = 3

So,

T(2) = 3+4 = 7

T(50) = 3 + (50-1)*4 = 3 + 196 = 199 .

Hence, our AP series will be :----

3, 7, 11 , 15, 19 , 23 , _________________ 199 .