Question

find the sum of 7, 11, 15, 19,....up to 60 terms​

Answer 1

Answer:

7500

Step-by-step explanation:

There are 2 methods to find the answer

Method 1:

S = n/2 { 2a + (n-1)d } where,

S is the sum, n is the number of terms, a is the first term and d is the common difference

s is unknown, n=60, a=7, d= 11-7 = 4

S = 60/2 {2 x 7 + (60-1) x 4}

  = 30 {14 + (59 x 4) }

  = 30 { 14 + 236}

  = 30 x 250

  = 7500

Method 2:

S= n/2 (a + l)

S is the sum, n is the number of terms, a is the first term and l is the last term.

Since we don't know what the last term is we should first find the last term

To find the last term we will apply T= a + (n-1)d

T = a + (n-1) d = 7 + (60-1) x 4 = 7 + 236 = 243

Now that we know the last term we can apply it to S= n/2 (a + l)

S= n/2 (a + l) = 60/2 x (7 + 243) = 30 x 250 = 7500

I hope this explanation is clear. But if you are going to use method 2, after finding the last term, remember to substitute it to S= n/2 (a + l). When following method 2, you might forget to substitute the l to the formula ( I myself and made this mistake many times in term tests) so  I would recommend you to for method 1 as it gives you the answer directly.

Answer 2

Given,

$\[7,11,15,19, - - - - 60terms\]$

Solution,

Know that the given series is arithmetic progression. its first term (a) is 7 and common difference (d) is 4. The number of terms n are 60.

Sum of n terms,

$\[\begin{array}{l}{S_n} = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\\ \Rightarrow {S_n} = \frac{{60}}{2}\left( {2 \times 7 + \left( {60 - 1} \right) \times 4} \right)\\ \Rightarrow {S_n} = 30 \times \left( {14 + 59 \times 4} \right)\\ \Rightarrow {S_n} = 30 \times 250\\ \Rightarrow {S_n} = 7500\end{array}\]$

Hence the sum of 60 terms is $\[7500\]$