The sum of first 7 terms …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Pulkit Upreti 4 years, 9 months ago
- 1 answers
Related Questions
Posted by Mohammad Ramjan 11 hours ago
- 1 answers
Posted by Kishan Kishan 3 days, 5 hours ago
- 2 answers
Posted by Chinmay M.R 3 hours ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 4 years, 9 months ago
Sn= {tex}\frac n2{/tex}[2a + (n - 1)d]
S7 = {tex}\frac 72{/tex}(2a + 6d) = 49
or, {tex}a + 3d = 7{/tex}...........(i)
S17 = {tex}\frac{17}{2}{/tex}(2a + 16d) = 289
or, {tex}a + 8d = 17{/tex}..........(ii)
On subtracting (i) from (ii), we get
or, 5d = 10 or, d = 2
Put d = 2 in (i)
a + 3d = 7
a + 2(3) = 7
a + 6 = 7
and a = 1
{tex}S _ { n } = \frac { n } { 2 } [ 2 \times 1 + ( n - 1 ) 2 ]{/tex}
{tex}= \frac { n } { 2 } [ 2 + 2 n - 2 ] {/tex}
{tex}= \frac { n } { 2 } [ 2 n ] {/tex}
{tex}= n ^ { 2 }{/tex}
Hence, sum of n terms = n2
0Thank You