Question

1. Find the sum of the following APs:
(1) 2,7,12, ..., to 10 terms.​

Answer 1

Answer:

Sum of n terms (Sn) = n/2{ 2a + (n -1)d }

(1) a = 2 , d = 7 - 2 = 5 and n = 10

S10 = 10/2{ 2×2 + (10 -1)× 5}

= 5 { 4 + 45}

= 5 × 49

= 245

(ii) a = -37 , d = 4 and n = 12

S12 = 12/2{ -37×2 + (12-1)×4}

= 6{ -74 + 44}

= -180

(iii) a = 0.6 , d = 1.1 and n = 100

S100 = 100/2{ 2 × 0.6 + (100-1)×1.1 }

= 50 × { 1.2 + 108.9}

= 50 × 110.1

= 5505

(iv) a = 1/15 , d = 1/12 - 1/15 = 1/60

S11 = 11/2{ 2/15 + 10 × 1/60 }

= 11/2 { 18/60 }

= 33/20

Step-by-step explanation:

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Answer 2

Answer:

a=2

common difference=7-2=5

sum(upto n terms)=n/2(2a+(n-1)d)

sum(upto 10 terms)=10/2(2×2+(10-1)5)

=5(4+45)

=5×50=250