Question

An A.P having 50 terms, in which if the sum of first ten terms is 210 and the sum of last fifteen terms is 2565 find ap​

Answer 1

Solution :-

Sum of first 10 terms = S₁₀ = 210

Using sum of n terms of an AP formula

Sₙ = n/2 (2a + (n - 1)d )

⇒ S₁₀ = 10/2 (2a + (10 - 1)d )

⇒ 210 * 2/10 = 2a + 9d

⇒ 42 = 2a + 9d --- eq(1)

Sum of last 15 terms S₁₅ = 2565

Sum of n terms of an AP = Sₙ = n/2(a + l)

Sum of last 15 terms of an AP = S₁₅ = 15/2(a₃₆ + a₅₀)

[ Because here, First term a = a₃₆ and last term l = a₅₀ ]

⇒ 2565 = 15/2 (a + 35d + a + 49d)

⇒ 2565 * 2/15 = 2a + 84d

⇒ 342 = 2a + 84d --- eq(1)

Subtracting equation (1) from equation (2)

⇒ 342 - 42 = 2a + 84d - 2a - 9d

⇒ 300 = 75d

⇒ 300/75 = d

⇒ 4 = d

Substituting d = 4 in eq(1)

⇒ 2a + 9d = 42

⇒ 2a + 9(4) = 42

⇒ 2a + 36 = 42

⇒ 2a = 42 - 36

⇒ 2a = 6

⇒ a = 6/2 = 3

General form of AP : a, a + d, a + 2d, a + 3d,....

Therefore the AP is 3, 7, 11, 15,....

Answer 2

Answer:

${\underline{\boxed{\sf\green{AP = 3,7, 11, 15,19,...........}}}}$

Step-by-step explanation:

$\frak{Given:-}\begin{cases}\sf\green{A.P. \: has = 50 \: terms}\\\sf\red{Sum \: of \: 1st\:10\:terms = 210}\\\sf\blue{Sum\: of \: last \: 15 \: terms = 2565}\\\sf\purple{A.P. = ?}\end{cases}$

At first,

Sum of 1st 10 terms = 210

We know sum of nth term of an AP,

${\underline{\boxed{\sf\blue{{S}_{n} = \dfrac{n}{2}[2a + (n-1)d] }}}}$

$\Rightarrow\sf {S}_{10} = \dfrac{10}{2}[2a + (10 - 1)d] \\\\\Rightarrow\sf 210 = 5 [2a + 9d] \\\\\Rightarrow\sf 2a + 9d =\cancel{\dfrac{210}{5}} \\\\\Rightarrow\sf 2a + 9d = 42 .........(i)$

$\rule{200}{1}$

Now,sum of last 15 terms = 2565.

We know formula for sum of nth term of an AP having the last few terms,

${\underline{\boxed{\sf\green{{S}_{n} =\dfrac{n}{2} (a + L) }}}}$

Last 15 terms among 50 terms = (36 - 50)

$\sf Where, a = {a}_{36} \: \& \: L ={a}_{50}$

$\Rightarrow\sf {S}_{15} = \dfrac{15}{2} ({a}_{36} + {a}_{50}) \\\\\Rightarrow\sf 2565 = \dfrac{15}{2} [a + (36-1)d + a +(50-1)d] \\\\\Rightarrow\sf 2565 = \dfrac{15}{2} (a + 35d + a + 49d) \\\\\Rightarrow\sf 2565 \times 2 = 15(2a + 84d) \\\\\Rightarrow\sf 5130 = 15(2a + 84d) \\\\\Rightarrow\sf 2a + 84d =\cancel{\dfrac{5130}{15}} \\\\\Rightarrow\sf 2a + 84d = 342........(ii)$

Now, by substracting (ii) from (i):-

$\Rightarrow\sf \cancel{2a} + 9d - \cancel{2a} - 84d = 42 - 342 \\\\\Rightarrow\sf -7 5d = -300 \\\\\Rightarrow\sf d =\cancel{\dfrac{-300}{-75}} \\\\\Rightarrow{\boxed{\sf\blue{ d = 4 }}}$

$\rule{300}{1}$

Putting value of d in (ii) we get:-

$\Rightarrow\sf 2a + 84\times 4 = 342 \\\\\Rightarrow\sf 2a + 336 = 342 \\\\\Rightarrow\sf 2a = 342 - 336 \\\\\Rightarrow\sf 2a = 6 \\\\\Rightarrow\sf a =\cancel{\dfrac{6}{2}} \\\\\Rightarrow{\boxed{\sf\green{ a = 3}}}$

We know terms of an AP :-

$\sf \green{a, a + d , a + 2d , a + 3d,.......}$

${\underline{\boxed{\therefore{\sf\purple{AP = 3, 7, 11, 15,............}}}}}$