25. Find the sum of 1, 3.5, 6, 8.5, …. Upto 21
terms.
a) 546
b) 520
c) 496
d) none​

2 thoughts on “25. Find the sum of 1, 3.5, 6, 8.5, …. Upto 21<br />terms.<br />a) 546<br />b) 520<br />c) 496<br />d) none​”

1. Given

   ⇒First term (a) = 1

   ⇒Common Difference (d) = a₂-a₁ = 3.5 – 1 = 2.5

   ⇒Number of term = (n) = 21

   Formula

   ⇒Sₙ = n/2{2a+(n-1)d}

   now put the value on formula

   ⇒S₂₁ = 21/2{2×1+ (21-1)2.5}

   ⇒S₂₁ = 10.5{2 + 20×2.5}

   ⇒S₂₁ = 10.5{2+ 50}

   ⇒S₂₁ = 10.5{52}

   ⇒S₂₁ = 546

   Answer

   ⇒S₂₁ = 546 , option ‘a’ is correct

   More Information

   ⇒a,b,c are in AP ; 2b = a+c , GP ; b²=ac and HP ; b = (2ac)/(a+c)

   ⇒AM = (a+b)/2 , G= √(ab) and H = (2ab)/(a+b)

   ⇒A≥G≥H In Equality Based

   Reply

2. Given :-

   An AP 1,3.5,6,8.5 Upto 21 terms

   To Find :-

   The sum

   Solution :-

   Finding common difference

   [tex]\sf C.D. = a_2 – a_1[/tex]

   CD = 3.5 – 1

   CD = 2.5

   We know that

   Sₙ = n/2{2a+(n-1)d}

   [tex]\sf S_{21}=\dfrac{21}{2} \bigg(2 \times 1 + (21 – 1 )2.5\bigg)[/tex]

   [tex]\sf S_{21}= \dfrac{21}{2} \bigg(2 + (21 – 1) 2.5\bigg)[/tex]

   [tex]\sf S_{21} = \dfrac{21}{2} \bigg(2 + 20 \times 2.5\bigg)[/tex]

   [tex]\sf S_{21} = \dfrac{21}{2}\bigg(2 + 50\bigg)[/tex]

   [tex]\sf S_{21} = \dfrac{21}{2}\times 52[/tex]

   [tex]\sf S_{21} = 21 \times 26[/tex]

   [tex]\sf S_{21} = 546 \bigg\lgroup Option \; A \bigg\rgroup[/tex]

   Reply