Question

Help me in this maths question pals!


Find the sum of first 20 terms in an A.P whose third term is 7 and 7th term is 2 more than thrice of 3rd trrm​

Answer 1

$<b> <u> <i> Heya!! </b> </u> </i>$

$<b> <u> Arithmetic Progressions</b> </u>$

✔Given that ,

→ a3 = 7

→$<b>a + 2d = 7..............(Eq.1)</b>$

Also ,

→ a7 = 3 (a3 ) + 2

→ a7 = 3 ( 7 ) + 2

→ a7 = 23

→$<b>a + 6d = 23.............(Eq.2)</b>$

★solving equations ( 1 ) & ( 2 ) we have ,

→$<b><u>d=4 </u></b>$

→$<b><u>a=-1 </u></b>$

★ Sum of first 20 terms :- S20

$= > \frac{n}{2} (2a + (n - 1)d) \\ \\ = > \frac{20}{2} (2 \times - 1 + (19) \times 4) \\ \\ = > 10( - 2 + 76)$

= 740

•°• $<b><u> Sum of first 20 terms is 740!!</u></b>$

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Answer 2

20th term =a+19d

3rd term = a+2d = 3

7th term = a+6d =23

so, a+2b =3

a+6d=23

........... -4d=-20

so, d=5

Sn = n/2 [2a+(n-1)d]

now put the value and solve the question