Question

determine the sum of first 35 terms of an a.p .. if second terms is 2 and the seventh term is 22?​

Answer 1

Answer

$\boxed{\boxed{\mathsf{{S}_{35} = 2310}}}$

Step-by-step Explanation :

Given

• Second term of an AP = 2

• Seventh term = 22

To find

• Sum of first 35 terms.

Solution

Now According to question

$\longrightarrow \mathsf{A.P = a_1 , (a_2 = 2 ), a_3 ........ (a_7 = 22)....}$

Now we know that,

$\longrightarrow \mathsf{a_2 = a_1 + d}$

$\longrightarrow \mathsf{a_3 = a_1 + 2d}$

Therefore,

$\longrightarrow \mathsf{a_7 = a_1 + 6d}$

Now putting value of a2 and a7 as per given question,

$\longrightarrow \mathsf{a_1 + d = 2}$ ---------- (1)

$\longrightarrow \mathsf{a_1+ 6d = 22}$ ---------(2)

Now Subtracting EQ 1 from 2 we have

$\longrightarrow \mathsf{a_1+ 6d - (a_1 +d) = 22 - 2}$

$\longrightarrow \mathsf{a_1+ 6d - a_1 - d = 22 - 2}$

$\longrightarrow \mathsf{5d = 20}$

$\longrightarrow \mathsf{d = \dfrac{20}{5}}$

$\longrightarrow \boxed{\mathsf{d = 4}}$

Now putting value of d = 4 in EQ (1)

$\longrightarrow \mathsf{a_1 + d = 2}$

$\longrightarrow \mathsf{a_1 + 4 = 2}$

$\longrightarrow \mathsf{a_1 = 2 - 4}$

$\longrightarrow \boxed{\mathsf{a_1 = -2}}$

Now to find Sum we have formula

$\boxed{\boxed{\mathtt{\bigstar \: S_n = \dfrac{n}{2} \bigg( 2(a_1) + (n - 1)d \bigg)}}}$

Where, here

$\mathsf{S_n = Sum\: of\: terms \: till \: n = {S}_{35}}$

$\mathsf{n =Number\: of \: terms = 35 }$

$\mathsf{d = common difference = 4}$

$\mathsf{a_1 = First term = -2}$

Now putting this values in the formula we have

$\longrightarrow \mathsf{{S}_{35} = \dfrac{35}{2} \bigg( 2(-2) + (35 - 1)4 \bigg)}$

$\longrightarrow \mathsf{{S}_{35} = \dfrac{35}{2} \bigg( (-4) + (34)4 \bigg)}$

$\longrightarrow \mathsf{{S}_{35} = \dfrac{35}{2} \bigg( (-4) + 136\bigg)}$

$\longrightarrow \mathsf{{S}_{35} = \dfrac{35}{2} \bigg( 132 \bigg)}$

$\longrightarrow \mathsf{{S}_{35} = 35 (66)}$

$\longrightarrow \boxed{\mathsf{{S}_{35} = 2310}}$

Therefore we have Required answer

$\underline{\boxed{\mathsf{{S}_{35} = 2310}}}$

Answer 2

Given :------

• 2nd Term of AP = 2
• 7th Term of AP = 22

To Find :------

• Sum of 35 terms of AP ..

Formula used :-----

nth term of AP is ,

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

sum of nth term of AP is,

$\large\red{\boxed{\sf Sn = \frac{n}{2}[2a + (n - 1)d]}}$

______________________________

Solution ,

putting values now in Tn formula we get,

T2 = a +d = 2 ----------------------- Equation (1)

T7 = a + 6d = 22 ----------------- Equation (2)

Subtracting Equation (1) from Equation (2) , we get,

(a+6d) - (a+d) = 22 - 2

→ 5d = 20

→ d = 4

putting in any Equation , now we get,

a = 2 - 4 = (-2)

so, our AP series is :--- -2 , 2, 6 , 10 _____________

Now , sum of First 35 terms of AP will be :------

Again , putting Values in above formula we get,

$Sn = \frac{35}{2} (2 \times ( - 2) + (35 - 1)4) \\ \\ Sn = \frac{35}{2} ( - 4 + 136) \\ \\ Sn = \frac{35 \times \cancel 132}{ \cancel2} \\ \\ Sn \: = 2310$

So, sum of 35 terms of AP will be 2310...

(Hope it Helps you)