
If the nth term of an arithmetic progression is\[3n + 7\], then what is the sum of its first \[50\] terms?
A) \[3925\]
B) \[4100\]
C) \[4175\]
D) \[8200\]
Answer
232.8k+ views
Hint: in this question we have to find sum of first \[50\] term of given AP series. nth term of AP is given in general form use this to find the first and second term then by using first and second term try to find the common difference of an AP. Then apply the formula of sum of AP to get required value.
Formula Used: In order to calculate sum of n terms of AP formula is given as:
\[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]
Where
\[{S_n}\] is sum of of AP
A is first term of AP
n is number of terms in AP
d is common difference of AP
Complete step by step solution: Given: \[3n + 7\]
Now \[{A_n} = 3n + 7\]
First term is \[{A_1} = a = 3 + 7 = 10\]
Second term is \[{A_2} = 3 \times 2 + 7 = 13\]
Common difference is given as \[{A_2} - {A_1}\]
\[d = {A_2} - {A_1}\]
\[d = 13 - 10 = 3\]
Now sum of series is given as
\[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]
\[{S_n} = \dfrac{{50}}{2}[2 \times 10 + (50 - 1) \times 3]\]
\[{S_n} = 25[20 + 49 \times 3]\]
\[{S_n} = 4175\]
Option ‘C’ is correct
Note: Here we must remember that we can find any term if general equation of n term of AP is known.
Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any pattern then follow that pattern to get required values. Sometimes by using pattern we are able to find the first term and common ratio or common difference therefore always try to find first term and common ratio or common difference if required. Then apply the formula to get the required value.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
Formula Used: In order to calculate sum of n terms of AP formula is given as:
\[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]
Where
\[{S_n}\] is sum of of AP
A is first term of AP
n is number of terms in AP
d is common difference of AP
Complete step by step solution: Given: \[3n + 7\]
Now \[{A_n} = 3n + 7\]
First term is \[{A_1} = a = 3 + 7 = 10\]
Second term is \[{A_2} = 3 \times 2 + 7 = 13\]
Common difference is given as \[{A_2} - {A_1}\]
\[d = {A_2} - {A_1}\]
\[d = 13 - 10 = 3\]
Now sum of series is given as
\[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]
\[{S_n} = \dfrac{{50}}{2}[2 \times 10 + (50 - 1) \times 3]\]
\[{S_n} = 25[20 + 49 \times 3]\]
\[{S_n} = 4175\]
Option ‘C’ is correct
Note: Here we must remember that we can find any term if general equation of n term of AP is known.
Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any pattern then follow that pattern to get required values. Sometimes by using pattern we are able to find the first term and common ratio or common difference therefore always try to find first term and common ratio or common difference if required. Then apply the formula to get the required value.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
Recently Updated Pages
Geometry of Complex Numbers Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

