Question

please answer this question​

Answer 1

Solution :-

Given That, Sum of first 10 Terms of AP is 210.

Let,

=> First Term = a

=> Common Difference = d

=> S(10) = (n/2)[2a + (n - 1)d]

Putting values we get,

=> 210 = (10/2)[2a + (10-1)d]

=> 210 = 5[2a + 9d]

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

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Now, we have also Given That, Sum of Last 15 Terms is 2565.

So,

=> First term now = 36th Term.

=> Tn = a + (n -1)d .

=> T(36) = a + (36-1)d

=> T(36) = a + 35d .

Putting values ,

=> 2565 = (15/2)[2(a + 35d) + (15-1)d]

=> 15 / 2 [ 2(a + 35d) + 14d ] = 2565

=> 15 [ a + 35d + 7d ] = 2565

=> a + 42d = 171 ---------------------- Equation (2)

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Multiply equation (2) by 2 and Than, Subtracting Equation (1) from This we get :-

=> 2(a + 42d) - (2a + 9d) = 2*171 - 42

=> 2a - 2a + 84d - 9d = 342 - 42

=> 75d = 300

=> d = 4.

Putting This value in Equation(1) we get,

=> 2a + 9*4 = 42

=> 2a = 42 - 36

=> 2a = 6

=> a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.

Answer 2

Given :

• An AP having 50 terms, n = 50.
• Sum of the first 10 terms of AP is 210.
• Sum of the last 15 terms of AP is 2565.

To Find :

• The terms in AP.

Solution :

Let the first term of the AP be a.

Let the common difference of the AP be d.

We know sum of n terms of an AP is,

$\large{\boxed{\bold{S_n\:=\:\dfrac{n}{2}\:\Big[2a\:+\:(n-1)\:d\Big]}}}$

For first 10 terms of AP :

Here,

• $\sf{S_n\:=\:210}$
• $\sf{n\:=\:10}$

Block in the data,

$\sf{\longrightarrow{210\:=\:\dfrac{10}{2}\:\Big[2a\:+\:(10-1)\:d\Big]}}$

$\sf{\longrightarrow{210\:=\:5\:\Big[2a\:+\:(9)d\Big]}}$

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

$\sf{210\:=\:10a\:+\:45d\:\:\:\:(1)}$

Now, sum of last 15 terms of the AP is 2565.

For last 15 terms of AP :

$\sf{\longrightarrow{15th\:term\:from\:last\:=\:(50\:-\:15\:+\:1)th\:term}}$

$\sf{\longrightarrow{15th\:term\:from\:last\:=\:(35+1)th\:term}}$

$\sf{\longrightarrow{15th\:term\:from\:last\:=\:36th\:term}}$

We know nth term of an AP is,

$\large{\sf{\red{t_n\:=\:a\:+\:(n-1)d}}}$

36th term of AP :

$\sf{\longrightarrow{t_{36}\:=\:a\:+\:(36-1)d}}$

$\sf{\longrightarrow{t_{36}\:=\:a\:+\:(35)d}}$

$\sf{\longrightarrow{t_{36}\:=\:a\:+\:35d\:\:\:(2)}}$

Sum of last 15 terms is 2565 :

$\sf{\longrightarrow{2565\:=\:\dfrac{15}{2}\:\Big[2a_{36}\:+\:(15-1)d\Big]}}$

$\sf{\longrightarrow{2565\:=\:7.5\:\Big[2\:(a+35d)\:+\:(14)d\Big]}}$

$\sf{\longrightarrow{2565\:=\:7.5\:\:(2a\:+\:70d\:+\:14d)}}$

$\sf{\longrightarrow{2565\:=\:7.5\:(2a\:+\:84d)}}$

$\sf{\longrightarrow{2565\:=\:15a\:+\:630d}}$

$\sf{\longrightarrow{2565-15a\:=630d}}$

$\sf{\longrightarrow{\dfrac{2565-15a}{630}=d\:\:\:(3)}}$

Now, consider equation (1),

$\sf{\longrightarrow{210=10a+45d}}$

$\sf{\longrightarrow{210=10a+45\Big(\dfrac{2565-15a}{630}\Big)}}$

$\bold{\Big[From\:equation\:(3)\:d\:=\:\dfrac{2565-15a}{630}\Big]}$

$\sf{\longrightarrow{210\:=\:10a\:+\:\Big(\dfrac{115425\:-\:675a}{630}\Big)}}$

$\sf{\longrightarrow{210\:=\:10a+\dfrac{115425-675a}{630}}}$

$\sf{\longrightarrow{210\:=\:\dfrac{6300a+115425-675a}{630}}}$

$\sf{\longrightarrow{210\:\times\:630\:=\:6300a\:+\:115425-675a}}$

$\sf{\longrightarrow{132300=6300a+115425-675a}}$

$\sf{\longrightarrow{132300-115425=6300a-675a}}$

$\sf{\longrightarrow{16875=5625a}}$

$\sf{\longrightarrow{\dfrac{16875}{5625}\:=\:a}}$

$\sf{\longrightarrow{3=a}}$

Substitute, a = 3 in equation (1),

$\sf{\longrightarrow{210\:=\:10a+45d}}$

$\sf{\longrightarrow{210=10(3)+45d}}$

$\sf{\longrightarrow{210=30+45d}}$

$\sf{\longrightarrow{210-30=45d}}$

$\sf{\longrightarrow{180=45d}}$

$\sf{\longrightarrow{\dfrac{180}{45}=d}}$

$\sf{\longrightarrow{d=4}}$

$\large{\boxed{\bold{First\:term\:=\:a\:=\:3}}}$

$\large{\boxed{\bold{Second\:term\:=\:t_2\:=\:a+d=3+4=7}}}$

$\large{\boxed{\bold{Third\:term\:=\:t_3\:=\:t_2+d=7+4=11}}}$

$\large{\boxed{\bold{Fourth\:term\:=\:t_4\:=\:t_3+d=11+4=15}}}$

$\large{\boxed{\bold{t_{50}\:=\:3+(49)(4)\:=\:3+196=199}}}$

$\large{\boxed{\bold{\purple{AP\:=\:3,7,11,15,.....199}}}}$